Question

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 34.0 m above sea level, directed at an angle θ above the horizontal with an unknown speed v0. The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 152.0 m.

1) Assuming that air friction can be neglected, calculate the value of the angle θ. You know how long the rock is in the air. This yields its horizontal velocity from the horizontal distance it travels. You get the vertical component of the velocity by using the equation of motion with constant acceleration. From the two velocity components you can figure out the angle.

2) Calculate the speed at which the rock is launched. 26.56m/s You know the net change in height of the rock, along with the horizontal distance it travels. Since you calculated the initial angle already, you can use your equations of position in two dimensions to solve for the initial speed.

3)To what height above sea level does the rock rise?

Answer 1

(a)

first we find the initial horizontal component of velocity

v_ox = D / t = 152 / 6 = 25.333 m/s

and

to find v_oy , we can use kinematics

-34 = v_oy * 6 - 1/2 * 9.8 * 6²

v_oy = 23.733 m/s

so,

= tan ^-1 ( 23.733 / 25.33)

= 43.13 degree

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(b)

v_o = v_ox / cos

v_o = 25.333 / cos 43.13

v_o = 34.71 m/s

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(c)

h_max = 23.733² / 2 * 9.8

h_max = 28.738 m

but we need max height from sea level

so,

h_max = 28.738 + 34

h_max = 62.738 m