Height of the object is considered to describe the motion when the body’s motion is in the vertical direction. In that case height conveys to the distance to find the velocity.

**Suppose an object is thrown into the air; the object used moves at a certain height and covers a certain distance. In that case, the body is making both vertical and horizontal motions, in that case how to calculate velocity with height and distance? The answer is discussed in this post.**

**How to Find Velocity with Height and Distance?**

When an object is projected upward, it reaches a certain height, and it falls back to the surface by covering a certain distance. The velocity of the object is to be calculated using these two data.

**The velocity of the projected object while traveling upward is different from the same object traveling down. Since the object travels in the vertically upward direction, and after reaching the maximum height, it again travels back on the ground, the height is not equal to the distance, as discussed in the earlier article.**

Since the object has traveled in both upward and downward directions, the total distance covered by the object is given by

x = d +h

Since the object is traveling back to the ground at the maximum height h, the peak where the object increases its velocity to travel back to the ground is given as,

As the object begins to fall from the peak, the height decreases, hence the total distance traveled by the object can be rewritten as,

Therefore, the total distance covered by the object can be expressed as,

We know that from the kinematic equation of motion, the distance traveled by the body moving the velocity v and begins to accelerate for every t seconds is given by;

We can substitute the otained value of x in the equation to get velcoity from height and distance as, to get

Rearranging the terms to get the velocity as

The above equation gives the answer for how to find velocity with height and distance.

**How to find initial velocity with height and distance?**

We have already discussed several ways to find the initial velocity. An accelerating body has various values of velocity. The initial velocity can be calculated using acceleration, but can we calculate the initial velocity with height and distance?

**The initial velocity of the body is calculated when the time t=0. At the time t=0, the height at which the body is and the distance, that the body has covered are used to find the velocity. The accelerating body keeps on changing the velocity; this helps us to find the initial velocity.**

Let us consider an example; a body is pushed from a certain height ‘h,’ and it begins to fall at a certain angle and travels at a certain Distance towards down. As the body is moving in the downward direction, its velocity becomes maximum due to gravitational pull.

In this case, the distance is along the horizontal plane, and height is along the vertical plane.

First, we deal with the height of the object because height is an entity that is influenced by gravity. The equation for the vertical displacement, i.e., the height of the body traveled, is given by

Here, we use acceleration due to gravity because, in the vertical direction, the body can accelerate only due to gravity.

In the vertical motion, we consider v=0, as initially, the body possesses zero velocity.

Hence by this equation, we can find the time interval t.

We can find the initial velocity in the horizontal motion because the body can also accelerate due to the velocity factor.

Now we have the time interval to cover the distance d. the initial velocity can be calculated as

The concept of finding initial velocity can be clearly understood by solving problems, which is done in the following section.

**How to find horizontal velocity with distance and height?**

Generally, we divide the components into vertical and horizontal motion while dealing with velocity problems. The displacement of the body in a horizontal direction over a given time period gives the horizontal velocity.

**The height can influence the velocity only when the body is traveling in the vertical direction. If the body is making both vertical and horizontal motion, then we need magnitude of the velocity to calculate horizontal velocity. Let us know how to find velocity with height considering the horizontal distance traveled by the body.**

The horizontal component of velocity is given by

v_{x}= v cosθ

The vertical component of velocity is given by

v_{y} = vsinθ

Since gravity is influencing over the vertical motion of the body, so the vertical components can be expressed as,

v_{y} = vsinθ-gt

The velocity can be expressed as

v = v_{x} + v_{y}

The magnitude of the velocity is expressed as,

If the body reaches maximum height, the vertical velocity v_{y} is zero. It can be given as

v_{y} = 0 = vsinθ-gt

gt = v sinθ

The total flight time of the body is given by 2t, therefore using the above expression, the time can be given as

g(t) = v sinθ

t=v sinθ/g

Let us assume that the range of the body under motion is nothing but the distance traveled by the body; it can be given as

d=v^{2} sin2θ/g

The height reached by the body while moving vertically is given by

h=(1/2)gt^{2}

Substituting the value of time interval t we get,

We can equate the obtained equation of height and the distance to get value of θ as

But we have to find the horizontal velocity, v_{x}= vcosθ

By using the height and distance equation, we can calculate the value of v, and hence substituting the value of v in the horizontal component of velocity equation we get the horizontal velocity.

**Solved example problems on how to find velocity with height and distance**

**Find the velocity of the projectile accelerating due to gravity, and is projected upward and reaches a maximum height of 12m and travels a distance of 42 meters. The time taken by the projectile to cover the distance is 1.33 seconds.**

**Solution:**

Given –maximum height reached by the projectile h = 12m.

Distance traveled by the projectile d = 42m.

Time taken by the projectile to travel given distance t = 1.33 sec.

Acceleration due to gravity g = 9.8 m/s^{2}.

The velocity for given height and distance is given by

v = 11.27 -6.527

v = 4.743 m/s

**A stone is thrown horizontally from the top of the hill at the height of 12 meters and the distance traveled by the stone in a horizontal direction is 23 meters. If the stone accelerates due to gravity, how to find velocity with height and distance?**

**Solution:**

Given – the height at which the stone is thrown h = 12m.

Horizontal distance traveled by the stone d = 23m.

Acceleration due to gravity g = 9.8m/s^{2}.

Since the stone is thrown horizontally, the vertical velocity is zero initially. The time taken to cover the given distance is given by

To find velocity, the general expression is given by

v=d/t

v=23/1.56

v=14.74m/s.

**Find the horizontal velocity of an object projected vertically at the height of 6 meters and cover the horizontal distance of 17 meters. (Take acceleration due to gravity is 10 m/s**^{2})

^{2})

**Solution:**

Given –the horizontal distance covered by the body d= 17m.

The vertical height h = 6m.

The horizontal velocity is given by

v_{x} = v cosθ

The value of theta for the horizontal velocity is given by tan^{-1}(2)=63.43

v^{2}=17/[10sin2(63.43)]

v^{2}=0.768

v=0.876 m/s.

**Find the initial velocity of the object accelerating with 5m/s**^{2} and the distance traveled by the object is 13 meters, and the height at which it is moving is 4 meters above the ground.

^{2}and the distance traveled by the object is 13 meters, and the height at which it is moving is 4 meters above the ground.

**Solution:**

Given –acceleration of the object is a = 5m/s^{2}.

Distance traveled by the object d = 13m.

Height of the object h = 4m.

Initially the vertical velocity of the object is zero and hence the equation can be written as

t^{2} = 1.6

t= 1.26 s.

The initial velocity can be given as,

v_{i}=d/t

v_{i}=13/1.26

v_{i} = 10.31 m/s.