An object is dropped from a height of 85.0m85.0m above ground level. Treat the effect of air resistance to be negligible and the upwards direction to be positive.

Question

 Include units as appropriate below.

a) Determine the distance travelled after 0.50s0.50s.

(b) Determine the velocity at which the object hits the ground.

(c) Determine the distance travelled in the last second of motion before the object hits the ground.

Answer 1

a) Determine the distance traveled after 0.50s

Let y = 0 at the location the object is released. In general, the distance the object has traveled t seconds after being dropped is:

y(t) = (1/2) × a × t²,

...so the location of the object is :

y(t = 0.50 s) = (1/2) × (-9.8 m/s²) × (0.50 s)² = -1.23 m

(b) Determine the velocity at which the object hits the ground.

The object hits the ground when y = -85 m at time t_g, so :

y(t_g) = (1/2) × (-9.8 m/s²) × (t_g²)= -85 m

Solving for t_g:

t_g = √( 2 × 85 /9.8) s = 4.165 s

In general, the velocity at time t is :

v(t) = v₀ + a × t

... so at impact the velocity is:

v(t_g) = -9.8 m/s × 4.165 s = -40.82 m/s

(c) Determine the distance traveled in the last second of motion before the object hits the ground.

One second before impact, the time elapsed is t_g - 1 = 3.165 s. The position at this time is

1/2 * (-9.8 m/s²) * (3.165 s)² = -49.08 m

The difference between that position and the position of the ground (85 m) is the distance traveled in the last second of motion:

85 m - 49.08 m = 35.92 m