Aball is thrown horizontally from the top of a building 0.10 km high. the ball strikes the ground at a point 65 m horizontally away from and below the point of release. what is the speed of the ball just before it strikes the ground?

The speed of the ball just before it strikes the ground is equal to 46.55 m/s.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 .

To determine the speed of the ball just before it strikes the ground:

First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.

Time, t = 4.52 seconds

Next, we would find the horizontal velocity:

Horizontal velocity, V1 = 14.38 m/s

Also, we would find the velocity of the ball in the horizontal direction:

Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:

Speed, V = 46.55 m/s

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v = 46.55 m/s

Explanation:

It is given that,

A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m

The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m

At maximum height, velocity of the ball is 0. So, using the equation of motion as :

Here, a = g

Let is the horizontal velocity of the ball. It is calculated as :

Let is the final speed of the ball in y direction. It can be calculated as :

Let v is the speed of the ball just before it strikes the ground. It is given by :

v = 46.55 m/s

So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.

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