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A baseball is thrown horizontally with a velocity of 7 m/s. It hits the ground 4.472 m

away. From what height was the ball released?

User Adam Lane
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1 Answer

5 votes

Answer:

2.002 m

Step-by-step explanation:

First, let's write out what we know about this problem and what we want to find.

We are given a horizontal velocity of 7 m/s

we are also given horizontal displacement = 4.472 m

For a projectile, we know ignoring air resistance

horizontal acceleration is 0 m/s^2

and Vertical acceleration is -9.81 m/s^2 the acceleration due to gravity

Now if this is only thrown horizontally then the initial vertical velocity =0

We are looking for the height the ball was released, which will have to do with the vertical displacement of the ball.

Before we solve for the vertical displacement, we must use the information in the horizontal to find time.

We will use the equation

displacement = Initial velocity (time) + 1/2 Acceleration * (time)^2

We are using horizontal dimension therefore all these values must be horizontal besides time because time is a scalar and does not have direction.

We get

4.472= 7 (t) + 1/2 (0)t^2 this will simplify to

4.472 =7 t

t=0.6389 s

Now we can use the same equation to solve for vertical displacement we will use all vertical quantities and time

Displacement = 0 m/s (0.6389) + 1/2 (-9.81)* (.06389)^2

This will give you a displacement of -2 m which means we will feel 2 meters down and must have started at a height of 2.002 m.

Hope this helps

User Michel Gammelgaard
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7.7k points