Question

CONSTRUCTION A roofer tosses a piece of roofing tile from a roof onto the ground 30 feet below. He tosses the tile with an initial downward velocity of 10 feet per second. a. Write an equation to find how long it takes the tile to hit the ground. Use the model for vertical motion, H = –16t2 + vt + h, where H is the height of an object after t seconds, v is the initial velocity, and h is the initial height. (Hint: Since the object is thrown down, the initial velocity is negative.) b. How long does it take the tile to hit the ground?

Answer 1

Given Information:

initial height = h = 30 ft

initial velocity = v = 100 ft/s

Required Information:

a) Equation of how long it takes the tile to hit the ground = ?

b) how long it takes the tile to hit the ground = ?

Answer:

a) $H = -16t^{2} -10t + 30$

b) $t = 1.09 \: s$

Explanation:

a) Write an equation to find how long it takes the tile to hit the ground. Use the model for vertical motion.

The equation for vertical motion is given by

$H = -16t^{2} +vt + h$

where H is the height of the object after t seconds, t is the time, v is the initial downward velocity, and h is the initial height.

Substituting the given information into the above equation yields,

$H = -16t^{2} +(-10)t + 30$

The negative sign is due to the fact that the object is thrown downward.

$H = -16t^{2} -10t + 30$

b) How long does it take the tile to hit the ground?

When the tile hits the ground, the height of the object is zero,

$0 = -16t^{2} -10t + 30\\\\0 = -2(8t^{2} + 5t - 15)\\\\0 = 8t^{2} + 5t - 15$

This is a quadratic equation so applying the quadratic formula yields,

$a = 8\\b = 5\\c = -15\\\\t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \\\\t = \frac{-5 \pm \sqrt{5^{2}-4(8)(-15)}}{2(8)} \\\\t = \frac{-5 \pm \sqrt{505}}{16} \\\\t = \frac{-5 \pm 22.47}{16} \\\\t = \frac{-5 + 22.47}{16} \\\\t = 1.09 \: s$

The other root is -1.71 and since time cannot be negative therefore, discard the negative root and accept the positive root that is 1.09 seconds.