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 39 m/s and an initial vertical velocity of 26 m/s. 1) What is the initial speed of the cannonball? 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? 4) How far from where it was shot will the cannonball land? 5) What is the speed of the cannonball 2.9 seconds after it was shot? 6) How high above the ground is the cannonball 2.9 seconds after it is shot?

Explanation / Answer

here,

initial horizontal velocity , ux = 39 m/s

initial vertical velocity , uy = 26 m/s

1)

initial velocity , u =sqrt(ux^2 + uy^2)

u = sqrt(39^2 + 26^2)

u = 46.87 m/s

2)

theta = arctan(uy/ux)

theta = arctan(26/39)

theta = 33.69 degree

the initial angle is 33.69 degree from the ground

3)

let the maximum height be h

h = uy^2 /( 2 * g)

h = 26^2 /( 2 * 9.8)

h = 34.49 m

4)

time of flight , T = 2 * uy /g

T = 2 * 26 /9.8

T = 5.31 s

the horizontal distance , x = ux * T

x = 39 * 5.31

x = 206.94 m

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