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6. A human cannonball at a circus is shot at an angle of 65.0" above the horizontal and at amspeed of 9.20If he leaves the cannon 2.00 m above the ground and lands in a net 4.00m above the ground, for what length of time is he in the air? (NOTE: He lands in the neton the way down.)

Question

I am poor at physics. Please explain as much as possible

 

6. A human cannonball at a circus is shot at an angle of 65.0" above the horizontal and at a
m
speed of 9.20
If he leaves the cannon 2.00 m above the ground and lands in a net 4.00
m above the ground, for what length of time is he in the air? (NOTE: He lands in the net
on the way down.)
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6. A human cannonball at a circus is shot at an angle of 65.0" above the horizontal and at a m speed of 9.20 If he leaves the cannon 2.00 m above the ground and lands in a net 4.00 m above the ground, for what length of time is he in the air? (NOTE: He lands in the net on the way down.)

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Step 1

To solve the question, the concept of projectile motion is used. The initial height and final height of the human projected through the cannonball is provided. 

Write the kinematic equation for vertical motion of the projectile.

1
-g2
h h ut-
(1)
2
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1 -g2 h h ut- (1) 2

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Step 2

Here, h is the human’s final height, ho is the human’s initial height, uy is the human’s initial velocity in the vertical direction and t is the time taken by the human to reach to a height during its motion.

Since, the human is projected at an angle of 65.0O at a speed 9.20 m/s, the initial vertical velocity is given by the sine component of the initial speed. That is,

u(9.20 m/s)sin 65.0°
=(9.20 m/s)0.906
и,
-8.34 m/s
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u(9.20 m/s)sin 65.0° =(9.20 m/s)0.906 и, -8.34 m/s

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Step 3

Substitute 8.34 m/s for uy, 4.0 m for h, 2.0 m for ho and 9.8 m/s2 for ...

(4.0 m)(2.0 m)+(8.33 m/s)t-(9.8 m/s2 )f
9.82-16.67t 4 0
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(4.0 m)(2.0 m)+(8.33 m/s)t-(9.8 m/s2 )f 9.82-16.67t 4 0

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Kinematics

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