Question

A person standing on the edge of a high cliff
throws a rock straight up with an initial
velocity of 17.50 m/s. The rock misses the edge
of the cliff as it falls back to Earth. What is the
position and velocity of the rock at 1.00 s, 3.5 s
and 4.0 s after it is thrown, neglecting the
effects of air resistance.

Answer 1

Given,

Initial velocity, u = 17.5 m/s

Now,

Upward direction is taken as positive

acceleration of the rock is in downward direction, a = -9.8 m/s²

Velocity at t = 1 s,

As we know,

v = u + at

=> v = 17.5 + (-9.8)*1 = 17.5 - 9.8

=> v = 7.7 m/s

Displacement, s = ut + 1/2*a*t²

=> s = 17.5*1 +(1/2)*(-9.8)*1²

=>    = 17.5 - (0.5*9.8)

=>    = 12.6 m

hence,

Velocity at 1 s is 7.7 m/s in upward direction

Position of the rock is 12.6 m above the cliff

Velocity at t = 3.5 s,

As we know,

v = u + at

=> v = 17.5 + (-9.8)*3.5 = 17.5 - (9.8*3.5)

=> v = -16.8 m/s

Displacement, s = ut + 1/2*a*t²

=> s = (17.5*3.5) +(1/2)*(-9.8)*(3.5)²

=>    = (17.5*3.5) - ( 0.5*9.8*12.25)

=>    = 1.225 m

hence,

Velocity at 3.5 s is 16.8 m/s in downward direction

Position of the rock is 1.225 m above the cliff

Velocity at t = 4 s,

As we know,

v = u + at

=> v = 17.5 + (-9.8)*4 = 17.5 - (9.8*4)

=> v = -21.7 m/s

Displacement, s = ut + 1/2*a*t²

=> s = (17.5*4) +(1/2)*(-9.8)*4²

=>    = (17.5*4) - (0.5*9.8*16)

=>    = -8.4 m

hence,

Velocity at 4 s is 21.7 m/s in downward direction

Position of the rock is 8.4 m below the cliff