Question

A rocket is launched with an initial velocity of 24.5 m/s from the ground. The height of the rocket over time is modeled by the function S(t)=− g/2× t^2 +v_0 t+h_0, where g is the gravitational acceleration, v_0 is the initial velocity, and h_0 is the initial height. After how many seconds will the rocket hit the ground?

A. 5 seconds
B. 6 seconds
C. 7 seconds
D. 8 seconds

Answer 1

The rocket will hit the ground after 5 seconds.

Step-by-step explanation:

To find the time at which the rocket hits the ground, we need to set the height function S(t) equal to zero and solve for t.

S(t) = -g/2 times t² + v0 times t + h0 = 0

Plugging in the values g = 9.8 m/s², v0 = 24.5 m/s, and h0 = 0 (since the rocket is launched from the ground), we have:

0 = -4.9t² + 24.5t

Now we can factor out t:

t(0 = -4.9t + 24.5)

Dividing both sides by t, we get:

0 = -4.9 + 24.5/t

Adding 4.9 to both sides:

4.9 = 24.5/t

Multiplying both sides by t:

4.9t = 24.5

Dividing both sides by 4.9:

t = 24.5 / 4.9

t = 5

Therefore, the rocket will hit the ground after 5 seconds.