Question

A ball is thrown into the air. Its height (in feet) t seconds later is given by:

Answer 1

Answer:

Step-by-step explanation:

Given that a ball is thrown into the air. Its height (in feet) t seconds later is given by:

$h(t)=128t-16t^2$

a) the height of the ball 1 second after it has been thrown

$= 128(1)-16(1) = 112$

b) The ball hits the ground when h =0

$h(t)=128t-16t^2$=0

t=0 or 8

After 8 seconds it hits the ground

c) $10 = 128t-16t^2\\8t^2-64t+10 =0\\t=\frac{64+/-\sqrt{64^2-320} }{16} \\=0.159 seconds, 7.841$

so after 0.159 seconds and 7.841 seconds it is at height 10 feet