Question

A particle is thrown with speed 10m/s vertically upward from a height of 10m from horizontal surface.Its average speed during time of flight is ?

Answer 1

u = 10m/s

S= 10m

Put g= 10m/s^2

Applying Equation of motion,

S=ut+(1/2)at^2

10=10t - 5t^2

t^2-2t-2=0

t^2-2t+t-2=0

t(t-2) + 1(t-2)=0

(t-2) (t+1) = 0

t=2 or t = -1

t cannot be negative so,

t = 2 sec

Average speed = Total Distance/Total Time

=10/2 = 5m/s

Answer 2

Answer:

Using equations of motion for free-fall :

v^2 =u^2 + 2ay subsitute to get u

100 = u^2 _ 294 , u = 19.8 m/s.

v= u + 9.8 t , 0 = 19.8 _ 9.8 t , t= 2.02 s time of moving up.

So , the time of flight is 2*t = 2*2.02.

Explanation: