A pebble is tossed from the top of a cliff. The pebble's height in feet is given by y(t) = -16t2 + 6t + 200, where t is the time in seconds. Its horizontal distance in feet from the base of the cliff is given by d(t) = 5t.
How far will the pebble be from the base of the cliff when it hits the ground? Round to the nearest foot. _______ ft Use a number only.

Question

A pebble is tossed from the top of a cliff. The pebble's height in feet is given by y(t) = -16t2 + 6t + 200, where t is the time in seconds. Its horizontal distance in feet from the base of the cliff is given by d(t) = 5t. 
How far will the pebble be from the base of the cliff when it hits the ground? Round to the nearest foot. _______ ft Use a number only.

campbell_st · Answer

the vertical distance will be zero so

y(t) = 0 

16t^2 -6t - 200 = 0

2(8t^2 - 3t - 100) = 0

using the general quadratic formula

$$t = (3 \pm \sqrt{3209})/16$$


only consider the positive t value

$$t = (3 + \sqrt{3209})/16$$

Substitute into the horizontal distance equation

$$d(t) = 5 \times (3 \pm \sqrt{3209})/16$$ 

the pebble will be 19 ft from the base of the cliff

anonymous · Answer

oh i get it thx