A ball is thrown up and is caught by the player after 4 seconds. How high did it go and with what velocity was it thrown? How far below the ball will be from its highest point after 3 second from start ?
Answers
Answer:
The maximum height reached by the ball is nothing but the displacement (s) of the ball
in terms of mechanics,
Time taken by the ball to go up and then come back again to be caught by the thrower = 4 secs
4
Time taken by the ball to reach its maximum height = - 2 sec
2
Acceleration due to gravity for this throwis given 'a' = 9.8 m/s
As the ball is being thrown upward, i.e in the opposite direction of the gravitational pull,
a'= -9.8 m/s2
Applying laws of kinematics to this event of ball reaching its maximum height,
v=u+at
Here v= 0 as on reaching maximum height, the ball stops moving.
..0=u-(9.8)2
Initial velocity of the ball thrown upward 'u'=19.6 m/s.
To calculate the maximum height i.e the displacement of the ball when reaching the highest point,
we use the following kinematical equation
1
S=ut+ -at-
s- at
s=19.6 (2)+L(-98) (2)
=
S=39.2-19.6= 19.6m
To calculate how far below its highest point was in 3 second after start:
The ball takes 2 secs to reach its maximum height i.e 19.6 m,
therefore the ball after 3 secs of throwing is 3-2-1 sec from the point of maximum height
towards the downward direction.
At one second from the point of maximum height,
u=0, displacement=s, time.
On applying the following laws of kinematics,
1
s=utt-at
S=
0 (1)+10.8) ay
1
S=4.9m
Therefore, the ball is 4.9 m away from the point of maxiumum height after 3 seconds of throwing the ball up
Explanation: