Example 2. A tennis ball is thrown up and
reaches a height of 4.05 m before coming
down. What was its initial velocity? How
much total time will it take to come down?
Assume g= 10 m/s2
Answers
Answer:
CASE 1: while throwing the ball upwards
Given : s = 4.05 m
a = -10 m/s2 (as the acceleartion is in the opposite direction)
v = 0 m/s (ball stops after reaching max height)
Equation = v2 - u2 = 2*a*s
(0)^2 - (u)^2 = 2*(-10)*4.05
-(u)^2 = 2*(-10)*4.05 - 0
-(u)^2 = -81
u = 9 m/s
Equation = v = u + a*t
0 = 9 + 10*t
-9 = 10 t
t = -9/10
= 0.9 seconds
CASE 2: the ball comes down
time of ascent = time of descent ( as they cover the same distance, and are under a constant value of free fall )
Adding both the times:
0.9 seconds + 0.9 seconds = 1.8 seconds
Therefore, the tennis ball will take 1.8 seconds to come down.
u = initial velocity
v = final velocity
a = acceleration due to gravity
t = time
s = distance or here, height
HOPE IT HELPS ;)