ng 1.pdf solving 1.pdf (96.2 KB) far up do they see the ball move (from when you

Question

ng 1.pdf solving 1.pdf (96.2 KB) far up do they see the ball move (from when you re the peak) and how far down (from the peak to when you catch i 7) Standing on a cliff, I take one rock and throw it straight up at a speed of 30 m/s. I take another rock and throw it straight down at 30 m/s. Suppose the cliff is 50 meters high. a) Just based on common sense, which rock would be moving faster when it hits the ground, 50 meters below? What's the reasoning for that? Now find an answer based on the kinematics of constant acceleration: Find x(t) and v(t) for each of the rocks, and find their respective speeds when they hit the ground 50 m below the point of release. b) HEWLETT-PACKARD ho 6 7 8 9

Explanation / Answer

7)

(a) Both rocks must hit the ground at same speed, as both will have a speed of 30 m/s downwards, when they go in downwards direction from the starting point.

(b) Case 1: Ball given initial velocity 30 m/s upwards.

As motion is governed by gravity in downward direction,

Hence,

v(t) = u + at

v(t) = 30 - gt

where downwards is negative direction and upwards is positive direction.

Also, distance covered will be

x(t) = ut + 1/2at2

x(t) = 30t - 0.5gt2

Using final distance x(t) = -50 and g = 10ms-2

-50 = 30t - 5t2

t2 - 6t -10 = 0

Which gives positive root of t = 7.36 s

Using this t, velocity will be

v = 30 - 10(7.36) = - 43.6 m/s

Case 2: when intial velocity 30 m/s in downward direction

u = -30 m/s

Hence,

v = -30 - gt

And

x(t) = -30t - 5t2

Using x(t) = -50

-50 = -30t - 5t2

t2 + 6t -10 = 0

Which gives positive value of t = 1.36 s

Using this t, velocity will be

V = -30 - 10(1.36) = -43.6 m/s

Which is exactly same as case 1

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