Question

A projectile is shot from the edge of a cliff 125 m above the ground. It has an initial velocity 50.0 m/s 37.0° above the horizontal at the launch point.
a) Determine the time taken to reach the ground. b) How far is this point from the base of the cliff? c) What is the magnitude and direction of the final velocity just before the point of impact

Answer 1

First, let's set up the equations of motion. We know that this is a projectile motion case. Let's begin by breaking the initial velocity (\(v_0\)) into its horizontal (x) and vertical (y) components. \[v_x = v_0 \cos(\theta)\]\[v_y = v_0 \sin(\theta)\]where \(\theta\) is the launch angle.

Now, let's realize that no forces act on the body in the horizontal direction. The horizontal displacement is expressed as\[x = v_x t ..... (1)\] and the horizontal velocity remains constant.
Let's realize that only gravity acts in the vertical direction. The vertical location is expressed as\[y = y_0 + v_y t - {1 \over 2} g t^2......(2)\]where \(y_0\) is the height of the cliff off of which the projectile is launched.
The vertical velocity is expressed as\[v_y(t) = v_y - g t......(3)\]

Now, we have all the equations we need.

(a) Find the time such that y in equation 2 is equal to zero.
(b) Plug this time back into equation 1 to find the horizontal distance.
(c) Plug the flight time into equation 3 to find the vertical velocity. The magnitude of the velocity is found as\[v = \sqrt{v_x^2 +v_y^2}\]and direction from\[\theta = \tan {v_y \over v_x}\]

Answer 2

exctly..