Projectile Motion

Question

A projectile is launched at ground level with an initial speed of 50.0 m/s and an elevation angle (i.e., at an angle above the horizontal) of 30.0°. It strikes a target above the ground 3.00 s later. What are the horizontal and vertical distances of the target from the place where the projectile was launched? Ignore air resistance.

Answer 1

For constant velocity (no acceleration) d = V*t

A) d horizontal = d_h = V_h*t because there's no horizontal acceleration = Vcos(30)*t = 50*cos(30)*3 = 129.9 m

Force = m*a and there's gravity which is a force in the vertical direction so there IS acceleration in the vertical direction so we must use different equation.

B)

(this comes from integrating acceleration*dt twice from calculus, but if u don't know calculus it's ok, just memorize this formula)

d = V₀*t + 0.5*a*t^2 (note: if a is zero then u just get d = V₀*t) in this case, gravity is accelerating in the opposite direction of motion (its pulling down but ur projectile is being shot up so you have to minus the acceleration). V₀ would be 50*sin(30) to get the vertical velocity and a = -9.8 m/s^2.

d = 50sin(30)*(3 seconds) - 4.9*((3 seconds)^2) = 30.9 m in the vertical direction.