A rock is shot into the air by a slingshot from ground level (y = 0). When the r

Question

A rock is shot into the air by a slingshot from ground level (y = 0). When the rock is 9.1 m high, its velocity is v = 7.6 m/s x + 6.1 m/s y. (a) What will be the maximum height of the rock above the ground? (b) What were the magnitude and direction of the initial (launch) velocity? (c) If the rock lands on top of a cliff with a height of 5.0 m, how far (horizontally) from the launch point will the rock land? (d) What are the magnitude and direction of the rock's velocity when it lands on the cliff?

Explanation / Answer

(A) at maximum component, y component of velocity will become zero.

Applying vfy^2 - v0y^2 = 2 a y

0^2 - 6.1^2 = 2 (-9.8)(y)

y = 1.90 m

height from ground, H = 1.90 + 9.1 =11 m

(B) x - component will always remains same.

v0x = 7.6 m/s

for y -c omponent: 6.1^2 - v0y^2 = 2(-9.8)(9.1)

v0y =14.7 m/s

magnitude = sqrt[ v0x^2 + voy^2]

v0 = 16.5 m/s .......Ans

directin angle = tan^-1(v0y / v0x) = 63 deg .....Ans

(C) 5 = 14.7t - 9.8 t^2 / 2

4.9 t^2 -14.7t + 5 = 0

t = 2.61 sec

x = 7.6 x 2.61 = 19.8 m

(d) vfx = 14.7 - (9.8 x 2.61) = -10.9 m/s

magnitude = sqrt(7.6^2 + 10.9^2) = 13.3 m/s

direction = tan^-1(10.9 / 7.6) = 54 deg below the horizontal.

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