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

ID: 1408077 • Letter: A

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 37 m/s and an initial vertical velocity of 22 m/s.

What is the initial speed of the cannonball?

I got for this one 43.05 m/s which was correct

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

30.7 degree . (correct)

What is the maximum height the cannonball goes above the ground?

24.7 m ( correct)

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

166.5 m

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5) What is the speed of the cannonball 3.4 seconds after it was shot?

How high above the ground is the cannonball 3.4 seconds after it is shot?

i could not get questions 5&6 right?

from the kinematic equation , the speed of the cannon in vertical direction is

vy= u+ at

=22- ( 9.8)(3.4)

=-11.32 m/s

horizontal veclocityof the canon is vx = 37 m/s

the speed of the cannon ball after 3.4 s

v = root vx^2 + vy^2 = ( 37 m/s)^2 +(- 11.32 m/s)^2 = 38.69 m/s

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(f)

the high above the ground is the cannonball 3.4 seconds after it is shot is

H=Vyt-(1/2)gt^2

=22 ( 3.4)-(1/2)*9.8 m/s^2*(3.4 s)^2

=18.156 m

H=18.1 m

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