Respuesta :
Answer:
0.93 seconds is the time interval between the release of the ball and the time it reaches its maximum height.
Explanation:
Given that,
Initial speed of the ball, u = 9.5 m/s
When it reaches to the maximum height, final speed of the ball, v = 0
It is required to find the time interval between the release of the ball and the time it reaches its maximum height. Using third equation of motion to find the maximum height reached.
[tex]v^2-u^2=2as\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(9.5)^2}{2\times 9.8}\\\\s=4.6\ m[/tex]
Now using second equation of motion to find the time interval between the release of the ball and the time it reaches its maximum height as :
[tex]s=ut+\dfrac{1}{2}at^2\\\\s=ut-\dfrac{1}{2}gt^2\\\\4.6=9.5t-\dfrac{1}{2}\times 9.8t^2\\\\t=0.93\ s[/tex]
So, 0.93 seconds is the time interval between the release of the ball and the time it reaches its maximum height.
Answer:
0.97 s
Explanation:
initial velocity of projection, u = 9.5 m/s
final velocity at maximum height, v = 0 m/s
acceleration due to gravity, g = 9.8 m/s²
Let the time taken is t.
Use first equation of motion
v = u - gt
0 = 9.5 - 9.8 x t
t = 0.97 s
Thus, the time elapsed as it reaches the maximum height is 0.97 s.
