Physics question

Question

A cannon situated on earth and pointing 53 degrees above horizontal, fires a cannonball from ground level at an initial velocity of 25 m/s.

a. what is the time it will take for the cannonball to hit the ground?

b. what is the maximum height the cannonball will reach?

Answer 1

A)

1. Calculate V in the vertical direction. V_v = V*sin(53 degrees) = 25*(4/5) = V_v = 20 m/s
2. D = V_v*t + 0.5*a*t^2, a = -9.8 m/s/s. Plug in and D = 20*t - 4.9*t^2. We want to find where the D = 0, because we're finding where the cannonball has gone up and come back down to the initial point. So we need to find the roots of this equation (the equation = D, so set the equation = 0, and solve for t).

0 = 20*t - 4.9*t^2 = t*(20 - 4.9*t) = 0, so this equals 0 at t = 0 (before it was shot) and also at t = (20/4.9) = 200/49 seconds = around 4.08 seconds. Note you could also have solved for maximum displacement, used that to find time and then doubled it (since it's the time to go up and then also back down, you would've gotten the same answer).

B)

1. Maximum height = D(total time/2) = D(100/49 seconds) = 20(100/49) - 4.9*(100/49)^2 = around 20.41 m.