The projectile remains in flight for 6.00 seconds and travels a horizontal dista
ID: 1706266 • Letter: T
Question
The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 164.0 m. Assuming that air friction can be neglected, calculate the value of the angle theta, the speed at which the rock is launched, and the maximum height above sea level the rock achieves.
A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H of 32.0 m above sea level, directed at an angle theta above the horizontal with an unknown speed V0 The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 164.0 m. Assuming that air friction can be neglected, calculate the value of the angle theta, the speed at which the rock is launched, and the maximum height above sea level the rock achieves.Explanation / Answer
for horizontal motion, it is a constant velocity motion
D = Vx*t (1)
Vx = D/t = 164 / 6 = 27.3 m/s
for vertical motion, it is under constant gravitational acceleration g
-H = Vyt -(1/2)gt^2 ( we choose the launch point as origin of y axis, so the final position y cordinate is -H)
Vy = ((1/2)gt^2-H)/t = (1/2)gt -H/t = 0.5*9.8*6 - 32/6 = 24.1 m/s
tan = Vy/Vx = 24.1/27.3 = 0.88
= 41.44 degree
V0 = sqrt(Vx^2+Vy^2) = sqrt(27.3^2+24.1^2) = 36.4 m/s
the maximum height reached when vertical velocity reaches 0 at time t
0 = Vy - gt
t = Vy/g
height above launch point
h = Vyt - (1/2)gt^2 = Vy(Vy/g) -(1/2)g(Vy/g)^2 = Vy^2/(2g)
= 24.1^2/(2*9.8) = 29.6 m
the maximum height above sea level = 29.6 + 32 = 61.6 m
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