how far will it go before it hits the ground?

Question

If a ball is launched by a slingshot at a 60° angle with a speed of 8.0m/s, how far will it go before it hits the ground?

Answer 1

You need to treat the horizontal and vertical velocities independently, and assume that air resistance is negligible.

Start by calculating the initial horizontal velocity v_x0 and the initial vertical velocity v_y0, and assign them both positive values. You know the intial velocity v_0 = 8 and the angle theta = 60. By trigonometry, v_x0 = v_0*cos(theta) and v_y0 = v_0*sin(theta).

Then use the gravity formula y = v_y0 + 0.5g*t^2 to find t for y = 0. This is because you need to figure out how much time elapses before the ball hits the ground (y = 0) at the end of its trajectory. You already calculated v_y0, and don't forget that g, gravity, is negative because we took the upwards direction to be positive. Finally, the horizontal distance traveled is just v_x0*t, because the horizontal velocity is constant due to ignoring air resistance.

Answer 2

First, you need to solve for the horizontal and vertical components of the initial velocities. So Vh = 8 m/s*cos 60= 4m/s; Vv = 8m/s*Sin 60 = 6.93 m/s. The problem is asking for the horizontal distance so it could be solved by the formula Dh = Vh*time (Horizontal distance =Horizontal velocity * total time of travel) Time could be solved by using the vertical velocity in finding the time of ascent then multiply it by 2 since time of ascent = time of descent. So, t = ((0 m/s -6.93 m/s) / 9.8m/s2 )*2= 1.41 s. Therefore multiplying the horizontal velocity of 4 m/s with the total time of 1.41 s will give a horizontal distance is 5.64 m.