Question

I solved the first 5 parts of the problem correctly but can’t seem to get part 6 please help me.

Cannonball A cannonball is shot (from ground level) with an initial horizontal velocity of 35 m/s and an initial vertical velocity of 29 m/s. 1) What is the initial speed of the cannonball? 45.45 Submit 2) What is the initial angle 0 of the cannonball with respect to the ground? 39.6 Submit 3) What is the maximum height the cannonball goes above the ground? 42.9 Submit 4) How far from where it was shot will the cannonball land? 207.14 m Submit

Answer 1

for 6

$y = (V_{i vert})t + \frac{1}{2}at^{2}$

here a = - g = - 9.8 m/s^2

$y = 29\times2.8 + \frac{1}{2}\times 9.8\times 2.8^{2} = 42.784 m$

so it will go 42.784 m above the ground