An angry mascot kicks a giant plastic egg filled with candy.

Question

The egg leaves the mascot's foot with a velocity of 8.4 m/s at an angle of 20 degrees above the horizontal. If you neglect air resistance, how far from the angry mascot will the egg land? Answer in meters. 

Answer 1

The answer is 4.63 meters.

We're being asked to find the egg's change in horizontal distance, Δx. The initial velocity is the sum of the initial velocity in the x direction and the initial velocity in the y direction, which we can find by representing the original velocity vector as the hypotenuse of a triangle with the y component being the vertical leg and the x competent being the horizontal leg. Therefore,

v₀ = v_x + v_y

v_x = v₀ · cos(20°), and v_y = v₀ · sin(20°)

Where v₀ is the initial velocity. Recall now that velocity times time equals displacement, so v_x · Δt = Δx. We know v_x = v · cos(20°) = 8.4 · cos(20°) m/s, but we don't yet know Δt. We can find it by thinking about what's happening in the y direction. Let's use the kinematic equation Δd = v₀ · t + (1/2) · a · t². Since the egg starts at ground level and hits the ground at the end of its flight, Δd = 0. a will be the acceleration due to gravity, since we assume no other force is acting on the egg. We then have

0 = v₀ · t + (1/2) · (-9.8 m/s²) · t²

Which, if we rearrange, gives us

t = -2 · v₀ / (-9.8 m/s²)

and since we're considering motion in the y direction, v₀ = v_y = v₀ · sin(20°). That means the time is

t = -2 · v₀ · sin(20°) / (-9.8 m/s²) = 0.586 seconds.

The reason we cared about time is so we could plug it back into the equation v_x · Δt = Δx. Doing so now gives us

Δx = 0.586 s · 8.4 m/s · cos(20°) = 4.63 m, our final answer.