A catapult on a cliff launches a large round rock towards a ship on the ocean be

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 35.0 m above sea level, directed at an angle theta above the horizontal with an unknown speed The projectile remains In flight for 6.00 seconds and travels a horizontal distance O of 172.0 m. Assuming that air friction can be neglected, calculate the value of the angle theta. 3.95x10* deg Calculate the speed at which the rock Is launched. 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. To what height above sea level does the rock rise?

Explanation / Answer

vo cos * t = 172     =>   v cos = 86/3 m/sec

t=6 sec

-35 = v0 sin *t - (5)*t*t

-35 = v sin (6) - 180

[180 -35 ]/6 = v sin

v sin =    145/6   m/sec

tan = 145*3/6*86 =   145/172  

=   40.13 degree    answer !!!!

v =   86/(3*cos)   = 28.666666/0.764584   =   37.4931   m/sec     answer !!!!

v*v - u*u = 2as

181.25 = 2(10)s

s = 9.0625    m

ht above sea level = 35 + 9.0625   = 44.0625 m

lifesaver !!!!

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