A rock is thrown from the top of a 19 m building at an angle of 53° above the ho

Question

A rock is thrown from the top of a 19 m building at an angle of 53° above the horizontal. (Ignore any effects due to air resistance.)

A rock is thrown from the top of a 19 m building at an angle of 53 above the horizontal. (Ignore any effects due to air resistance.) (a) If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown? X m/s Enter a number. (b) How long is it in the air? (c) What is the velocity of the rock just before it strikes the ground? m/s) m/s)

Explanation / Answer

here,

height of the building , h0 = 19 m

theta = 53 degree

a)

let the time taken to hit the ground be t and initial speed be u

h0 = - u * sin(theta) * t + 0.5 * g * t^2

19 = - u * sin(53) * t + 0.5 * 9.8 * t^2 ...(1)

and

for horizontal range

x = h0 = u * cos(theta) * t

19 = u * cos(53) * t ...(2)

from (1) and (2)

u= 10.5 m/s

t = 3 s

the initial speed of the ball is 3 s

b)

the time of flight is 3 s

c)

u = (u * cos(theta) i + u * sin(theta) j)

u = ( 10.5 * cos(53) i + 10.5 * sin(53) j)

u = (6.32 i + 8.39 j) m/s

accelration , a = - 9.8 j m/s^2

for t = 3 s

the final velocity , v = u + a * t

v = ( 6.32 i - 21.01 j) m/s

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