The ball is thrown vertically upwards from the height of 14.1 meters. It takes 3 seconds till the ball hits the ground. What is the initial speed of the ball?
Y = 14.1 + Vo*t -4.9*t^2
Set Y = 0 and t = 3 and solve for Vo, which is the initial vertical velocity.
Vo = [9*4.9 -14.1]/3
= 30/3 = 10.0 m/s
Thanks! :)
To find the initial speed of the ball, we can use the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
Where:
h = initial height (14.1 meters)
u = initial velocity (what we need to find)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken (3 seconds)
Plugging in the values, we get:
14.1 = u(3) + (1/2)(-9.8)(3)^2
To simplify, we calculate (1/2)(-9.8)(3)^2 = -44.1 meters
Now, the equation becomes:
14.1 = 3u - 44.1
To isolate u (initial velocity), we rearrange the equation:
3u = 14.1 + 44.1
3u = 58.2
Dividing both sides by 3, we get:
u = 58.2 / 3
u ≈ 19.4 m/s
Therefore, the initial speed of the ball is approximately 19.4 m/s.