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

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 34 m/s and an initial vertical velocity of 26 m/s. What is the initial speed of the cannonball? _________ m/s What is the initial angle theta of the cannonball with respect to the ground? __________ 3) What is the maximum height the cannonball goes above the ground? _________ m How far from where it was shot will the cannonball land? _____ m What is the speed of the cannonball 1.1 seconds after it was shot? _____ m/s How high above the ground is the cannonball 1.1 seconds after it is shot? _______ m

Explanation / Answer

Here ,

a) for the initial speed of the cannonball

initial speed of the cannonball = sqrt(vx^2 + vy^2)

initial speed of the cannonball = sqrt(34^2 + 26^2)

initial speed of the cannonball = 42.8 m/s

2)

initial angle of cannonball = arctan(vy/vx)

initial angle of cannonball = arctan(26/34)

initial angle of cannonball = 37.4 degree

3)

for the maximum height

maximum height = uy^2/(2 g)

maximum height = 26^2/(2 * 9.8)

maximum height = 34.5 m

4)

distance where the cannon ball land = ux * 2uy/g

distance where the cannon ball land = 34 * 2 * 26/9.8

distance where the cannon ball land = 180 m

5)

after 1.1 s

speed of cannonball = sqrt(34^2 + (26 - 9.8 * 1.1)^2)

speed of cannonball = 37.3 m/s

6)

height of the cannonball after 1.1 s = uy * t - 0.50 gt^2

height of the cannonball after 1.1 s = 26 * 1.1 - 0.50 * 9.8 * 1.1^2

height of the cannonball after 1.1 s = 22.7 m

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