Answer to Question #177260 in Financial Math for daniel wagaye

Question

An object is thrown vertically upward from a height of h₀ ft with an initial speed of v₀ ft/sec. Its height h (in feet) after t seconds is given by

h= -16t² + v₀t + h₀. Given this, if it is thrown vertically upward from the ground with an initial speed of 64 ft/sec,

(a) At what time will the height of the ball be 15 ft? (two answers)

(b) How long will it take for the ball to reach 63 ft?

Answer 1

(a)$15=-16t^{2}+64t$

$=16t^{2}-64t+15$

$t= \frac {-b± \sqrt{b^2-4ac}}{2a}$

$t=\frac{64±\sqrt{4096-960}}{2\times16}$

$t=\frac{64+56}{32}=3.75$

$t=\frac{64-56}{32}=0.25$

3.75 sec and 0.25 sec

(b)$63=-16tt^{2}+64t$

$16t^{2}-64t+63$

$t=\frac{64± \sqrt{4096-4032}}{2\times16}$

$t=\frac{64+8 }{32}=2.25$

$t=\frac{64-8}{32}=1.75$

2.25 sec

1.75 sec