Answer:
2.04 sec and 20.38 meters
Explanation:
upward speed of the ball u = 20 m/s
final velocity of the ball v= 0 m/s
we know that [tex]v^2-u^2= 2gh[/tex]
[tex]0^2- 20^2= 2\times 9.81\times h[/tex]
on solving we get
h= 20.38 meters. this the maximum height it will reach.
to calculate when it will reach maximum height
use u= gt
t= u/g
⇒ t= 20/9.81 = 2.04 sec after this much time it will reach to its maximum height.
Answer:
Time taken by the ball to reach maximum height = 2.041 seconds.
Maximum height that ball reached = 20.41 meters.
Explanation:
Given that the initial speed of the ball, [tex]\rm u = 20\ m/s.[/tex]
We know, when the ball moves in upward direction, the acceleration due to gravity acts on it in downwards direction, therefore the affecting acceleration that acts on the ball [tex]\rm a = -g[/tex].
g is the acceleration due to gravity having value [tex]\rm 9.8\ m/s^2[/tex].
Also, the final speed of the ball at its maximum height will be 0, i.e., [tex]\rm v=0\ m/s.[/tex]
To find the time taken by ball to reach maximum height.
Let it reaches the maximum height in time t, using the following relation of Kinematics,
[tex]\rm v=u+at\\0=20+(-9.8)t\\-20 = -9.8 t\\t=\dfrac{-20}{-9.8}=2.041\ s.[/tex]
To find the maximum height.
Let the maximum height upto ball goes be s, then using,
[tex]\rm v^2-u^2=2as\\s=\dfrac{v^2-u^2}{2a}\\=\dfrac{0^2-20^2}{2\times (-9.8)}\\=\dfrac{-400}{-19.6}\\=20.41\ m.[/tex]
