Question

A student throws a softball horizontally from a dorm window 15.0 m above the ground. Another student standing 10.0 m away catches the ball at a height of 1.50 m above the ground. What is the initial speed of the ball?

Answer 1

The initial speed of the softball can be calculated using the height it drops and the horizontal distance it travels. By determining the time of fall for the vertical drop and knowing the horizontal distance, the initial horizontal velocity can be found, approximately 6.02 m/s.

Step-by-step explanation:

To solve for the initial speed of the softball, we need to analyze the motion in two dimensions separately: horizontal and vertical.

Horizontal Motion

The horizontal motion can be considered with constant velocity since there's no acceleration in that direction (ignoring air resistance).

Vertical Motion

The vertical motion can be described by the kinematic equations for uniformly accelerated motion (free fall). Given the height difference from the window to the catchpoint (13.5 m) and the acceleration due to gravity (9.81 m/s²), we can calculate the time it takes for the ball to fall this height. The equation we use is:

h = vit + (1/2)at²

Substitute h = 13.5 m, a = 9.81 m/s², and vi = 0 m/s (since the ball is thrown horizontally, the initial vertical speed is 0) to solve for t.

We find that the time t is approximately 1.66 seconds. Using this time and the horizontal distance of 10.0 m, we can now calculate the initial horizontal speed.

vh = d/t

Substituting d = 10.0 m and t = 1.66 s, the initial horizontal speed vh is approximately 6.02 m/s.

Answer 2

Speed of ball equals 6.024 m/s.

Step-by-step explanation:

Let the student throw the ball with a velocity of 'v' m/s horizontally.

Now the time in which the ball travels 10.0 meter horizontally shall be equal to the time in which it travels (15.0-1.50) meters vertically

Hence the time taken to cover a vertical distance of 13.50 meters is obatined using 2 equation of kinematics as

`s=(1)/(2)gt^(2)\\\\t=\sqrt{(2s)/(g)}\\\\t=\sqrt{(2* 13.5)/(9.81)}\\\\\therefore t=1.66 seconds`

Since there is no acceleration in horizantal direction we infer that in time of 1.66 seconds the ball travels a distance of 10 meters

Hence the spped of throw is obatines as

`Speed=(Distance)/(Time)\\\\v=(10)/(1.66)=6.024m/s`