Calculus projectile Problem! NEED HELP!

Question

Suppose the acceleration due to gravity at the surface of the moon is a = -6 ft/sec^2. From an initial height of 240 ft above the surface of the moon, a projectile is shot vertically upward, with an initial velocity of 48 ft/sec.

1. Express v, the velocity in ft/sec as a function of time t in sec.
2. Express s, the position (height) in feet as a function of time t in sec.
3. When will the projectile reach its maximum height?
4. what is the maximum height?
5. when will the projectile strike the surface of the moon?

Answer 1

Hello Lane,

By supposing a(t) = -6, the anti-derivative yields the velocity function,

v(t) = -6t + C

Part 1: From the initial condition for velocity of v(0) = 48, we have

48 = -6(0) + C

→ C = 48

→ v(t) = -6t + 48

Part 2: The anti-derivative of velocity yields position

→ s(t) = -3t² + 48t + D.

With the initial condition of s(0) = 240, we get

s(0) = -3(0)² + 48(0) + D = 240

→ D = 240

→ s(t) = -3t² + 48t + 240

Part 3: The object will reach its maximum height when the velocity is zero, as that is the moment the object has come to a momentary stop, between going up and coming down. So

v(t) = -6t + 48 = 0

→ 6t = 48

→ t = 8 seconds

Part 4: To find the maximum height, plug the 8 seconds into the position function

→ s(8) = -3(8)² + 48(8) + 240

= -3(64) + 384 +240

= -192 +624

= 432 ft

Part 5: When the projectile hits the surface, the position will equal zero

→ s(t) = -3t² + 48t + 240 = 0

→ t² - 16t - 80 = 0

→ (t - 20)(t + 4) = 0

→ t = 20, -4

But t = -4 would not make sense (going back in time), so

t = 20 seconds.

Thank you for the question, and have a great evening!

Michael Ehlers

Answer 2

1) v(t) = at + v₀ = (-6 ft/s²)·t + 48 ft/s

2) s(t) = (1/2)at² + v₀t + s₀ = (1/2)(-6)t² + 48t + 240 = -3t² + 48t + 240

3) When v(t) = 0. Solve equation 1 for t when v(t) = 0.

4) Plug the value of t you computed in (3) and plug it into s(t) in (2). The units are feet.

5) Set s(t) = 0 and solve for t. Factor or use the quadratic formula. You'll get two answers, one positive and one negative. Since we can't go back in time just yet, the positive value is the answer (units are seconds)