Question

A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude of the ballast after t seconds, (b) the time when it strikes the ground, and (c) its velocity when it strikes the ground. (Disregard air resistance and take ft/sec2.)

Answer 1

Answer:

a) $S = \frac{1}{2}gt^2$

b) 6secs

c) 192ft

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

$S = ut + \frac{1}{2}at^{2}$

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

$S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2$

a) An expression for the altitude of the ballast after t seconds is therefore

$S = \frac{1}{2}gt^2$

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

$576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs$

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft