Physics Homework

Question

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a second snowball at a low angle and time it to arrive at the same time as the first. Assume both snowballs are thrown with the same initial speed 29.8 m/s. The first snowball is thrown at an angle of 55◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? Note the starting and ending heights are the same. The acceleration of gravity is 9.8 m/s 2 . (part 2 of 2) How many seconds after the first snowball should you throw the second so that they arrive on target at the same time? Answer in units of s.

Answer 1

One range equation (derived) from t_air = 2v_y0/g and Range = v_x0t_air is Range = (2/g)v_x0v_y0

Knowing that v_x0 = v_ocos(θ₀) and v_y0 = v_osin(θ₀) , we obtain Range = (v₀²/g)sin(θ)cos(θ)

The range is going to be the same for θ and 90 - θ because sin(θ) = cos(90-θ) and vice versa.

That means that you hit the same place shooting at 55° and 35°. The time in the air for each angle is from the expression t_air = 2v_y0/g

I'll leave the calculation to you for the difference in the times.