You have an spring gun. The end of the gun is at 1 meter above the ground. This

Question

You have an spring gun. The end of the gun is at 1 meter above the ground. This gun is shoot horizontally with no air resistance. The cannon ball hits the ground 1.32 meters ahead from it was shoot. What is the cannon ball's initial velocity? Collect the data given by the problem (if the data is not given or implied leave it in blank): Initial position in the horizontal direction Initial position in the vertical direction Final position in the horizontal direction Final position in the vertical direction Initial velocity in the horizontal direction Final velocity in the horizontal direction Initial velocity in the vertical direction Final velocity in the vertical direction Initial time t_i = Final time Acceleration in the horizontal direction Acceleration in the vertical direction

Explanation / Answer

xi = 0 m

yi = 1m

xf = 1.32 m

yf = 0 m

Vyi = 0

ay = - 9.8 m/s^2

in vertical ,

yf - yi = Vyi*t + ay*t^2 /2

0 -1 = 0 - 9.8t^2 /2

t = 0.45 s

ti = 0

tf = 0.45 s

Vxi = (xf - xi)/t = 1.32/0.45 = 2.92 m/s

ax = 0

Vxf = Vxi + ax*t = 2.92 + 0 = 2.92 m/s

Vyf = 0 + (-9.8 x 0.45) = - 4.41 m/s

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