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

Question

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

Answer 1

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

Answer 2

I mean U = speed * cos angle up
Vi = speed * sin angle up

Answer 3

3.465

Answer 4

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

Answer 5

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

Answer 6

Nice move, thumbs up
Tnx

Answer 7

thanks you

Answer 8

Fabulous

Answer 9

Please the formula for calculating the maximum height

Answer 10

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