Question

A ball is thrown vertically upwards. It rises to a height of h m in 2.5 s and then

comes back to the thrower. Calculate

i) its initial speed,

ii) the time for which it remained in the air.

iii) the distance travelled and the displacement of the ball
​

Answer 1

Explanation:

as an example I have given you 2nd figure read it carefully and solve your problem

Answer 2

Answer:

i> Using formula v=u-gt where v(final velocity)=0

Taking g=10m/s

Therefore gt=u=>10m/s×2.5s=u=>u=25m/s(Ans)

ii>Time taken by the ball to raise up=time taken by the ball to come down=2.5s

Therefore the time for which it remained in the air=2.5s×2=5s(Ans)

iii>Using the formula v^2=u^2-2gS where v(final velocity)=0

u^2=2gS

=>(25)^2=2×10×S

=>625=20S

=>S=625/20=>31.25m

Therefore distance travelled=31.25m×2=62.5m

Displacement=0 [as the ball goes up and comes back to the same place]