A ball is thrown into the air Its height in feet t seconds l

A ball is thrown into the air. Its height (in feet) t seconds later is given by h(t) = 112t - 16t^2 Based on the two equivalent forms of the function h(t) = 112t - 16t^2 = t middot (112 - 16t), answer the following questions: What is the height of the ball 1 second after it has been thrown?___________ (include units in your answer) After how many seconds does the ball hit the ground? __________ (include units in your answer) At what time(s) is the ball 30 feet above the ground? If there is more than one answer, give your answer as a comma separated list of values. Note that you do not have to include units in this answer. After ____________ seconds

Solution

We are given that a ball is thrown upwards (means against the gravity of the earth) and the height or distance covered by the ball in \'t\' sec given by the equation => h = 112*t - 16*t²

a)

Height of the ball 1 sec after it is thrown is:

h = 112*1 – 16*1²
            h = 112 – 16
            or h = 96 feet

b)

After how many seconds the ball will hit back the ground, first we will find the time taken in covering maximum height :

Now, h(max) is when final velocity (v) = 0, but we don’t know the initial velocity (u) by which the ball is thrown, so to find ‘u’ we will use formula :

h = u*t - ½ *a*t²     [here acceleration \'a\' is -ve because direction of motion is upwards, against the gravity],      we have,

96 = u*1 - ½ *32*1²    [acceleration = gravity of earth = 32 ft/sec²] [result used from part (a)]
96 = u – 16
or u = 112 ft/sec

Now to find time taken in covering maximum height when final velocity (v) = 0, using the formula,

v = u – a*t we get
0 = 112 – 32*t
or 32*t = 112, gives t = 112/32 = 3.5 sec

> Time taken by the ball in achieving maximum height is 3.5 sec, and we know that the same time will be consumed by the ball in coming back on the earth, so the ball will hit back the ground after 2*t sec = 7 sec.

c)

The ball will be at 30 ft above the ground at two times, first when going upwards and second when coming downwards to the earth.

So, using the given relation => h = 112*t - 16*t²

We have, 30 = 112*t - 16*t²  
or   8*t² – 56*t + 15 = 0, which is a quadratic equation in variable \'t\'

Now by using the quadratic formula, t = { -b ± (b² - 4ac)} / 2a

We get, time (t) = 6.721 sec & 0.279 sec [with a = 8, b = -56, c = 15]

So, at 0.279 sec, 6.721 sec after throwing the ball, it is at 30 ft above the ground.