Question

16. A ball projected into air with velocity 20m/sec making an angle 45 with the horizontal
Find a) Range
b) Time of Flight 5​

Answer 1

Answer:

Given Data :

u=20m/s

θ=45

To Find :

R

solution :

on resolving velocity we get

uV=usinθ

uH=ucosθ

where uV : vertical component of initial velocity

uH : horizontal component of initial velocity

Now we know that horizontal acceleration is 0 therefore only acceleration on a motion is vertical acceleration which is g : acceleration due to gravity

Now we have to find time at which vertical velocity become 0

to find vV=uV+aVt

but aV=−g

and vV=0

∴0=uV−gt

⟹t=uV / g

⟹t=20sinθ / 9.8

⟹t=209.8–√ 2

but we know that path followed by projectile is symmetric \therefore total time period (T)=2t

⟹T=409.8–√ 2

Now to find horizontal Range R=uHt+1/ 2aHT2

but aH=0

∴R=uH40 / 9.8–√2

∴R=20∗40 / 9.8–√2–√ 2

⟹R=400 / 9.8

⟹R=40.817m

hope it will help you.....