physics question

Question

A cannon is placed on the edge of a 300m high cliff as shown in the figure. The cannon is fired at angle of  = 30 with respect to the horizontal and the cannonball leaves the cannon at a speed of 30m/s. a) What is the horizontal distance from the cannon to the impact site? b) What is the velocity of the cannonball just before it hits the ground?

Answer 1

In the y-direction the time it takes for the cannonball to hit the ground is the same as if the cannonball were dropped from the 300 m cliff. So:

h = 1/2gt²

300 = (1/2)(9.81)t²

t² = 61.16

t = 7.82 seconds

We can also calculate the velocity it has (in the y-direction) when it hits the ground from v = gt = (9.81)(7.82) = 76.72 m/s

In the x-direction, the cannonball leaves the starting location at a velocity of 30•cos(30°) = 25.981 m/s and it travels for the 7.82 seconds calculated above. Therefore the distance traveled is (25.981)(7.82) = 203 meters.

To calculate the velocity of the cannonball when it hits the ground, we must use both the velocity in the x-direction (25.981 m/s) and the velocity in the y-direction (76.72 m/s). To combine these, use the Pythagorean Theorem to get the magnitude: v = √(v_x² + v_y²). To get the direction, use inverse tangent: θ = tan^-1(v_y/v_x)