A group of shark engineers have created a cannon that can launch a typical great

Question

A group of shark engineers have created a cannon that can launch a typical great white shark from sea level at angle of 30 degree above the horizontal. Like most engineers, they crave coffee. As a result they would like to use this cannon to reach a starbucks located on the beach a little way inland (also at sea level;). If the starbucks is located 300 m from the cannon, with what speed should the sharks be launched? After the sharks become a nuisance at the starbucks, the city of TooManySharksville has had enough. They decide to build a Wall between the starbucks and cannon, but they only have enough funding to built the wall 30 m high. If they build the wall at the halfway point between the cannon and the starbucks, will it stop the sharks? If not, how much higher will the sharks be than the wall when they pass over it? Where should the city build the wall in order to stop the sharks? (To stop the sharks, assume height of the wall = height of the shark. Give answer in terms of the distance from starbucks. There are tow possibilities, you may choose either one.)

Explanation / Answer

here ,

theta = 30 degree

a)

distance ,d = 300 m

let the speed is v

range = v^2 * sin(2 * theta)/g

300 = v^2 * sin(2 * 30)/9.8

solving for v

v = 58.26 m/s

the initial speed must be 58.26 m/s

b)

maximum height = v^2 * sin^2(theta)/(2 * g)

maximum height = 58.26^2 * sin^2(30)/(2 * 9.8)

maximum height = 43.3 m

the shark will go over the wall

extra wall height needed = 43.3 - 30

extra wall height needed = 13.3 m

c)

let the distance is x m from starbucks

y = x * tan(theta) - g * x^2/(2 * (v * cos(theta))^2)

30 = (300 - x) * tan(30 degree) - 9.8 * (300 -x)^2/(2 * (58.26 * cos(30))^2)

solving for x

x = 66.9 m , 233.1 m

the location of wall must be 66.9 m and 233.1 m

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