Question

A ball is launched upward at 20 meters per second (m/s) from a 60 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 20t + 60, where s is in meters.
What is the height above the ground when the object is launched?
How long before the object hits the ground after launch?
What is the maximum height of the object?

Answer 1

we are given

$s(t)=-4.9t^2+20t+60$

(a)

we can plug t=0

$s(0)=-4.9(0)^2+20(0)+60$

$s(0)=60$

so, the height above the ground when the object is launched is 60 meter......Answer

(b)

we can set s(t)=0

and then we can solve for t

$s(t)=-4.9t^2+20t+60=0$

$-4.9t^2+20t+60=0$

we can use quadratic formula

$t=\frac{-200-\sqrt{200^2-4\left(-49\right)600}}{2\left(-49\right)}$

we get

$t=6.0917seconds$..........Answer

(c)

we can find vertex

$t=\frac{-20}{2*-4.9}$

t=2.040816

now, we can plug it back

$s(2.040816)=-4.9(2.040816)^2+20(2.040816)+60$

$s(2.040816)=80.408$

so, the maximum height is 80.408meter.........Answer