Question

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Answer 1

Answer:

t=5s

Explanation:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at

2

=20(2)+0.5(−10)(2)

2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at

2

45=0+0.5(10)(t

′

)

2

t

′

=3s

Total time= 3+2= 5s

pls mark brainliest

Answer 2

Given :-

A  ball is thrown vertically upwards with a velocity of 20m/ s from the top of a multi-story building.The height of the point where the ball is thrown 25 m from the ground

To Find :-

Time taken to hit the ground

Solution :-

We know that

v = u + gt

0 = 20 + (-10)t

0 - 20 = -10t

-20 = -10t

20 = 10t

20/10 = t

2 = t

Now

s = ut + 1/2 × gt²

s = 20(2) + 1/2 × (-10) × (2)²

s = 40 + (-5) × 4

s = 40 - 20

s = 20 m

Now

Total height = 20 + 25 = 45 m

s = ut + 1/2 × gt²

45 = 0(t) + 1/2 × (10)(t²)

45 = 0 + 5t²

45 = 5t²

45/5 = t²

9 = t²

√9 = t

±3 = t

As time can't be negative. t = 3

Hence

Total time = 3 + 2

Total = 5 sec