A cannon is situated 14m high above the ground and shoots a projectile at 20m/s. What is the location of the projectile?

Answer 1

Shooting directly right:

Initially there is only velocity in the x direction (20 m/s). That x-direction velocity doesn't change as the cannon moves because there are no forces in the x direction acting on the cannonball. But there is a force acting on it in the y direction - gravity. The cannonball, in the y-direction, acts just like it would if it were dropped from the edge of the cliff. The kinematic equation y_f - y_i = v_it + 1/2at² can be used. The initial velocity in the y-direction is zero (the only velocity initially is in the x direction), and y_f - y_i = -14, and a is the acceleration due to gravity which is -9.81 m/s² so -14 = 1/2(-9.81)t² or t = √(14•2)/9.81 = √2.8542 = 1.689 s

Since the cannonball, in the x-direction, is always going 20 m/s, to find the distance it goes, multiply 20 m/s times the 1.689 s to get 33.8 m. If you only have 1 significant figure, you would need to round to 1 sig fig or distance = 30 m but I'm not sure if you've really got 1 sig fig or if you just typed the problem like that.

Answer 2

Assuming it is launched horizontally at 20 m/s from a height of 14 m, get the time-of-flight from the height, and then multiply the time-of-flight by the horizontal speed. That will give the horizontal distance traveled when it hits the ground.

h = gt²/2 ⇒

t = √(2h/g)

Plug in

h = 14 m

g = 9.8 m/s²

and get t in seconds. Then the horizontal distance is

x = vt

Plug in

v = 20 m/s

t = result from above