A watermelon is sot out of a catapult with a velocity of 37m/s at an angle of 17

Question

A watermelon is sot out of a catapult with a velocity of 37m/s at an angle of 17.9 degrees above the horizontal. When the watermelon is released from the catapult it is 8.75m above the ground. Assume that air resistance is negligible and that the ground does not change in elevation. Calculate the time that the watermelon is in the air, calculate the horizontal and vertical components of the final velocity of the watermelon just before it strikes the ground, and if we substitute a 500 pound bomb for the watermelon, which of the answers, if any, would change (Do not recalculate them, just state which kinematics variables would have different values, if any)

Explanation / Answer

it have to go vertically downwards 8.75 m .

in vertical using

dy = uy*t + ay*t^2 /2

-8.75 = (37sin17.9)*t - 9.8t^2 /2

4.91t^2 - 11.37t - 8.75 = 0

t =2.92 sec

there is no acceleration in horizontal direction.

so, horizontal component will not change.

vx =ux = 37cos17.9 =35.02 m/s

vy = uy + ayt

vy = 37sin17.9 - 9.8 x 2.92 = -17.25 m/s

in any of equation, mass is not used.

so whatever object is used , all values will be same.

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