Question #68f57

1 Answer
Sep 28, 2017

1) Velocity after 2 second
v=10.4 m/s

2) Maximum Height Attained
s=45.91 m

3) Time Taken to reach the Maximum height
t=3.06 sec

4) Time taken to the ball to return to the ground
T=6.06 sec

Explanation:

Initial Velocity u=30 m/s
For upward motion Acceleration due to gravity a=-g=-9.8 m/s^2
1) Velocity after 2 second
t=2 sec
Using Newton's First Equation of Motion
v=u+at
v is the final velocity after t seconds.
v=30-g t=30-(9.8)(2)=30-19.6=10.4 m/s
v=10.4 m/s

2) Maximum Height Attained
On the Maximum height final velocity of object will be zero.
v=0
Using Newton's Third Equation of Motion
v^2=u^2+2as
s is distance traveled
=>0=30^2-2(9.8)s
19.6s=900
s=900/19.6=45.91 m

3) Time Taken to reach the Maximum height
On the maximum height final Velocity v=0
Using Newton's First Equation of Motion
v=u+at
v is the final velocity after t seconds.
0=30-g t=30-(9.8)(t)
9.8t=30
t=30/9.8
t=3.06 sec

4) Time taken to the ball to return to the ground
The time taken to reach maximum height and return to the ground is the same.
Hence Total time =2xxtime taken to reach the maximum height
T=2xx3.06=6.06 sec