Projectile Motion Problem

Question

A pumpkin is launched from the ground at an angle of 60° above the horizontal. The pumpkin strikes the center of a large target that is 80 ft from the pumpkin launcher and 48 ft above the ground. (g = 32 ft/s²)

1. What is the magnitude pumpkin’s initial velocity in [ft/s]?
2. What is the magnitude pumpkin’s velocity just before it strikes the target in [ft/s]?
3. What is the angle, in [^o], of the pumpkin’s velocity just before it strikes the target?

Answer 1

The first equation to use is y=v₀sin60^ot - 16t² where y=48ft. The second equation to use is x=v₀cos60^ot where x= 80ft. So 80=v₀t/2 and t=160/v₀. Hence t²= (160/v₀)² substitute these expressions into the equations to get

48=v₀(.867)160/v₀ -16(160/v₀)² and solve for v₀. Doing this gives v₀ =67.25ft/s. So t=160/v₀ =160/67.25=2.38s. To find the velocity v find v_y and v_x . So v_y = v_0y -gt so v_y= 67.25(.867) - 32(2.38) giving v_y = -18.397ft/s v_x = v₀(1/2) = 33.625ft/s. v=(v_x² +v_y²)^1/2 and this becomes v=75.187ft/s. The angle that the velocity vector makes with the target becomes θ=tan^-1(18.397/33.625) and θ= 28.68° below horizontal.