Question

A stone is projected at a cliff of height ℎ = 52 ? with an initial speed of 42.0 ?/? directed at angle ?? = 60.0° above the horizontal. A. Find the vertical and horizontal components of the ball’s initial velocity. (2 mark) B. Find the maximum height ? that the ball reaches. C. Find at what time the stone strikes ?. D. Find the ball velocity (in unit vector nutation) just before it strikes ?. E. At what horizontal distance from the edge of the building does the ball strikes ?? F. What is the vertical acceleration of the ball at the maximum height?

Answer 1

A)

v_x = 42 * cos 60 = 21 m/s

v_y = 42 * sin 60 = 36.4 m/s

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B)

at maximum height, v_y,f = 0

so,

H = 36.4² / 2 * 9.8

H = 67.5 meters ( above the ground)

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C)

we just want to find time in air

y_f = y_o + v_yt - 1/2gt²

52 = 0 + 36.4 t - 4.9t²

so,

4.9t² - 36.4t + 52 = 0

solve this equation

t = 36.4 +/- 17.48 / 9.8

t = 5.5 seconds

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d)

v_x = 21 m/s

v_y = 36.4 - 9.8 * 5.5 = -17.5 m/s

so,

v = 21 i - 17.5 j

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e)

x = v_xt

x = 116 m

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f)

a = g = -9.8 m/s²