A stone is thrown vertically upwards from a platform 60meters above the ground. The initial velocity of the stone is 20m/s.The height of the stone above the ground at t seconds is given by: H(t)=20t-5t^2+60m.How long does it take for the stone to hit the ground? At what velocity does the stone hit the ground?

what does it mean when it hits the ground? height H(t) = 0. so, since

H(t) = -5t^2 + 20t + 60
= -5(t^2 - 4 - 12)
= -5(t-6)(t+2)

when do you think it hits the ground?

Then, since

V(t) = 20 - 10t

you can figure the velocity

To find out how long it takes for the stone to hit the ground, we need to determine the time when the height of the stone (H(t)) becomes zero.

Given the equation for the height of the stone at time t is H(t) = 20t - 5t^2 + 60.

Setting H(t) to zero and rearranging the equation, we get:
0 = 20t - 5t^2 + 60

This equation is a quadratic equation in the form of at^2 + bt + c = 0, where:
a = -5
b = 20
c = 60

Using the quadratic formula, which is t = (-b ± √(b^2 - 4ac)) / 2a, we can find the values of t when H(t) = 0.

Substituting the values of a, b, and c into the quadratic formula, we have:
t = (-(20) ± √((20)^2 - 4(-5)(60))) / (2(-5))

Simplifying the equation further:
t = (-20 ± √(400 + 1200)) / -10
t = (-20 ± √1600) / -10
t = (-20 ± 40) / -10

This results in two possible time values:
t1 = (-20 + 40) / -10 = 2
t2 = (-20 - 40) / -10 = 6

Since time cannot be negative and we're interested in the time it takes for the stone to hit the ground, we discard t2 = 6.

Therefore, the stone takes 2 seconds to hit the ground.

To find out the velocity at which the stone hits the ground, we can differentiate the height equation with respect to time (t):

v(t) = dH(t)/dt = d/dt(20t - 5t^2 + 60)

Taking the derivative of each term:
v(t) = 20 - 10t

Substituting t = 2 into the equation:
v(2) = 20 - 10(2)
v(2) = 20 - 20
v(2) = 0

The velocity at which the stone hits the ground is 0 m/s.