A ball is thrown straight upwards with an initial velocity 19.6 ms–1. It was caughtat the same distance above the ground from which it was thrown:

A ball is thrown straight upwards with an initial velocity 19.6 ms–1. It was caughtat the same distance above the ground from which it was thrown:

(i) How high does the ball rise?

(ii) How long does the ball remain in air? (g = 9.8 ms–2)

Ans: – We know, v^2 = u^2 + 2as;

Here, v=final velocity=0; u= initial velocity= 19.5;

So, S= (19.5)^2 / 2×9.8=19.40m.

And, v= u + at,

Or, 0= 19.5 + 9.8×t;

Or, t = 1.98 or 2.

Then total time required is (2 + 2)= 4s.

 

A brick is thrown vertically upwards with the velocity of 192.08 ms–1to thelabourer at the height of 9.8 m. What are its velocity and acceleration when itreaches the labourer?

Ans: – As we all know the acceleration is same its just be in negative when it goes in upper direction so the acceleration will be -9.8m/s^2.

As the labour will catch the ball so it will have no velocity in that.

 

A body starts its motion with a speed of 10 ms–1 and accelerates for 10 s with10 ms–2. What will be the distance covered by the body in 10 s?

Ans: – We know, S= ut + 1/2at^2;

Here, u=10 m/s, a=10 m/s^2; t=10s.

So, S= 10×10 + ½× 10×10^2.

Or, S= 600m.

 

A car starts from rest and covers a distance of 50 m in 10 s and 100 m in next10 s. What is the average speed of the car?

Ans: – The average speed is = total distance/ total time.

So, average speed= (50 + 100)/20=150/20=7.5m/s.

Then the average speed of the car will be 7.5m/s.

 

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