Question

On a bright sunny day with no wind you set up a projectile canyon on the edge of a 300 m cliff. You launch with initial velocity of 18.5 m/s at an angle of 40 degrees.

How long before it hits the water?

How far away from the cliff will it land?

I need help setting this up and understanding how to solve it

Answer 1

here,

the height of cliff ,h = 300 m

initial velocity , u = 18.5 m/s

theta = 40 degree

let the time taken to hit the water be t

for vertical direction

- h = u * sin(theta) * t - 0.5 * g * t^2

- 300 = 18.5 * sin(40) * t - 0.5 * 9.81 * t^2

solving for t

t = 9.13 s

the time taken to hit the water is 9.13 s

the horizontal distance traveled , x = u * cos(theta) * t

x = 18.5 * cos(40) * 9.13 m

x = 129.4 m