Question

A ball is thrown upward with a certain speed. It passes through the same point at 3 second and 7 second from the start. The maximum height achieved by the ball is

1. 500 m

2. 250 m

3. 125 m

4. 450 m

Answer 1

Answer:

The maximum height achieved by the ball is 125 m

Explanation:

According to the problem the ball is thrown upward.

After 3 second it will reach at the top most position within = 7-3/2 = 2 sec

Therefore the ball will take 3+2 or 5 sec to reach at the top from the bottom.

Therefore for downward direction the motion will be,

u = 0, acceleration, a= g= 10 m/s^2,  t = 5 sec

Let the maximum height is h

hence h = 1/2 x g x t^2 = 1/2 x 10 x 5^2 = 125 m

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Answer 2

Answer:

The maximum height achieved by the ball is 125 m . option (3)

Explanation:

Here, the ball is thrown upward direction.

After 3 second it will reach at the top most position within = (7-3)/ 2 = 2 sec.

Therefore the ball will take (3+2) = 5 sec to reach at the top from the bottom.

For downward direction the motion will be,

u = 0, acceleration, a= g= 10 $m /s^2$,  t = 5 sec

Let the maximum height is h.

hence,

$h = \frac{1}{2} a t^2$= $= \frac{1}{2} * 10* (5)^2 =125$         [Using, $s = ut +\frac{1}{2} a t^2$   ]

∴ h = 125 meter