Question

A ball is thrown upwards vertically of 25m/s if g s 10m/s then calculate the height it reaches and time taken to return back

plz answer as fast as possible with step by step explanation
and plz no wrong answers plz....​

Answer 1

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Answer 2

★Given :-

• Initial velocity of the ball = 25 m/s
• Acceleration due to gravity = 10 m/s²
• final velocity of the ball = 0 m/s

(as velocity of the ball at highest point is always zero )

★ To find :-

• time taken by the ball to return back
• height reached

★ Solution :-

As the ball moves with uniform acceleration throughout it's motion we can use kinematic equations in order to solve this question

First let us find the highest point reached by the ball in order to do that simply apply third equation of motion

>> v² = u ² + 2gh

>> 0 = 625 + 2 x - 10 x h

(As the ball is moving against gravity acceleration will be negative )

>> h = 625 /20

>> h = 31.25 m

The highest point reached by the ball is 31.25 m

Now let's calculate the time of ascent i.e the time required by the ball to reach the maximum height in order to do that let's use the first kinematic equation ,

>> v = u + at

>> 0 = 25 -10 × t

>> t = 25 /10

>> t = 2.5 sec

• time of ascent = time of descent

hence ,

• time of ascent = time of descent = 2.5 sec

Total time taken by the ball to return back

= time of ascent + time of descent

= 2.5 + 2.5

= 5 sec

Total time taken by the ball to return back is 5 sec