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# PHYSICS EASY HELP ME PLEASE?

1.A car moving with a speed of 108 km/h is brought to rest in a distance of 60 m. How much time did the car take to stop?

2 seconds

4 seconds

6 seconds

8 seconds

2.A student standing on the ground next to a tall building throws a ball straight up with a speed of 39.2 m/s. If it takes the ball 4 seconds to reach the same height as the top of the building, how tall is the building?

a.44.1 m

b.78.4 m

c.161.7 m

d.235.2 m

3.A boy throws a ball with an initial velocity of 19.6 m/s and it returns to his hands in 4 seconds. What maximum height does the ball reach?

4.9 m

9.8 m

19.6 m

39.4 m

4.A man walks for 5 km/hr for 1 km and 7 km/hr for next 1 km. What is his average speed for the walk of 2 km?

5.8 km/hr

6.0 km/hr

2.0 km/hr

1.0 km/hr

### 2 Answers

- Wayne DeguManLv 71 month ago
2) a(t) => -g = -9.8

v(t) = -9.8t + v(0)

so, v(t) = -9.8t + 39.2

Then, s(t) = -4.9t² + 39.2t + s(0)

Assuming s(0) = 0 we have:

s(4) = -4.9(4)² + 39.2(4) => 78.4 metres

Note: In reality, when the ball is released from the hand it will be above the ground at a height of the student plus arm length...say about 2.5 metres

Hence, s(t) = -4.9t² + 39.2t + 2.5

When t = 4, this would give a height of 78.4 + 2.5

:)>

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- FiremanLv 71 month ago
Given u = 108 km/hr = (108 x 1000)/3600 m/s = 30 m/s

By v^2 = u^2 + 2as

=>0 = (30)^2 + 2 x a x 60

=>a = -7.5 m/s^2

Now by v = u + at

=>0 = 30 - 7.5 x t

=>t = 4 sec

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