A cannonball is shot (from ground level) with an initial horizontal velocity of

ID: 1591144 • Letter: A

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 36 m/s and an initial vertical velocity of 29 m/s.

1) What is the initial speed of the cannonball? m/s

2) What is the initial angle of the cannonball with respect to the ground? °

3) What is the maximum height the cannonball goes above the ground? m

4) How far from where it was shot will the cannonball land? m

5) What is the speed of the cannonball 3.3 seconds after it was shot? m/s

6) How high above the ground is the cannonball 3.3 seconds after it is shot? m

1) ux = 36 m/s

uy = 29 m/s

speed = sqrt(ux^2 + uy^2) = 46.23 m/s

2) @ = tan^-1(uy/ ux) = 38.85 deg

3) maximum height will be when vertical velocity is zero.

using vf^2 - vi^2 =2ad

0^2 - 29^2 = 2(-9.81)(h)

h = 42.86 m

4) time of flight t = 2vsin@ / g = 2 * 46.23 * sin38.85 / 9.81

t = 5.91 sec

x = ux*t = 36 * 5.91 =212.84 m

5)vx = ux = 36 m/s

vy = uy + ay*t = 29 - (9.81 * 3.3) = -3.37 m/s

speed = sqrt(vx^2 + vy^2) = 36.16 m/s

6) using h = uy*t + ay*t^2 /2

h = (29 * 3.3) + (-9.81 * 3.3^2 /2 )

h = 42.28 m

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