a ball dropped from rest at a height of 50 m above the ground. (a) what is the speed just before it hits the ground? (b) how long does it take to reach the ground?
a. -31
b . 3.16 sec
490
To find the answers to these questions, we can use the laws of motion and basic equations of kinematics.
(a) Finding the speed just before the ball hits the ground:
We can use the equation of motion that relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s):
v^2 = u^2 + 2as
In this case, the ball is initially at rest (u = 0), the acceleration due to gravity is acting in the downward direction (a = 9.8 m/s^2), and the displacement is equal to the height from which it is dropped (s = 50 m). Plugging in these values, we can solve for the final velocity:
v^2 = 0 + 2(9.8)(50)
v^2 = 980
v = √980
v ≈ 31.3 m/s
Therefore, the speed of the ball just before it hits the ground is approximately 31.3 m/s.
(b) Finding the time it takes to reach the ground:
We can use the equation of motion that relates initial velocity (u), acceleration (a), and time (t):
s = ut + (1/2)at^2
In this case, we know the initial velocity is zero (u = 0), the acceleration due to gravity is acting in the downward direction (a = 9.8 m/s^2), and the displacement is equal to the height from which it is dropped (s = 50 m). Plugging in these values, we can solve for the time:
50 = 0 + (1/2)(9.8)t^2
50 = 4.9t^2
t^2 = 50/4.9
t ≈ √(50/4.9)
t ≈ 3.19 s
Therefore, it takes approximately 3.19 seconds for the ball to reach the ground.
a. V^2 = Vo^2 + 2g*h
Vo = 0
g = 9.8 m/s^2
h = 50 m.
Solve for V.