PHYSICS PROBLEM I NEED HELP!!!!?

Question

A car is traveling at a constant speed of 28 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 1.2 km away?

AND

A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 10.7 m/s, releases a sandbag when the balloon is 39.9 m above the ground. Neglecting air resistance, what is the bag's speed when it hits the ground? Assume g = 9.80 m/s2.

Greatly appreciated!

Answer 1

For the first car travelling at constant speed u, the distancee s covered in time t is:

s = ut

t = s / u ...(1)

For the second car travelling with uniform acceleration a:

s = at^2 / 2

t^2 = 2s / a

Substituting for t^2 from (1):

s^2 / u^2 = 2s / a

2u^2 = as

a = 2u^2 / s

= 2 * 28^2 / 1200

= 1.31 m/s^2.

Let:

u be the initial upward speed of the balloon,

v is the speed at which the bag hits the ground,

s be the distance it falls.

v^2 = (- u)^2 + 2gs

= 10.7^2 + 2 * 9.80 * 39.9

v = 29.9 m/s.