A stone is projected at an angle of 60 degrees and an initial velocity of 20m/s calculate

(a) The time of flight
(b) The maximum height attained
(C) The time taken to reach maximum height
(d) Range

I mean U = speed * cos angle up

Vi = speed * sin angle up

3.465

Hey, you try first. If you get stuck come back but there is no point in my just doing it for you.

In general
split problem into vertical and horizontal parts
Horizontal problem:
U, horizontal speed, does not change
U = speed * sin angle up
so range = U T
Vertical problem:
Vi = speed initial * cos angle up
v = Vi - g t
at top v = 0
so at top t = Vi/g
H = Hi + Vi t - (1/2) g t^2
Hi is initial height, use top t to get height
Now drop from that height
v down = 0 + g t
at bottom
0 = Height max - (1/2) g t^2
solve for t down
then time in air T = t rising + t falling
range = U T

Well since we want to find the time of fight we can say t =2Usin60°/g our initial velocity is 20m/s

Then we substitute t= 2( 20)sin60°
And sin60° is 0.866
2×20 = 40 so 40×0.866 = 34.6s
So therefore t=34.6s

Nice move, thumbs up

Tnx

thanks you

Please the formula for calculating the maximum height

A stone is projected at an angle 60° and an initial velocity of 20 metre per second square. Determine the maximum horizontal distance covered

Caculate the time flight of a stone is projected at an angle of 60 degree and an initial velocity of 20m/s

Fabulous