Differential calculus, Physics problem

In summary, the projectile fired with a velocity of 400m/s reaches a maximum altitude of 2500m and hits the ground after 25 seconds at a velocity of 400m/s. The acceleration of the projectile at any time is -32m/s^2.
  • #1

Homework Statement

A projectile is fired straight upward with a velosict of 400m/s. From physics, its distance above the ground after t seconds is: s(t) = -16t^2 + 400t

A Find the time and velocity at which the projectile hits the ground

B Find the maximum altitude achieved by the projectile

C Find the acceleration at any time t.

Homework Equations

s(t) = -16t^2 + 400t

The Attempt at a Solution

A The surface s(t) = 0

0 = -16t^2+400t
16t^2-400 = 0
16t(t-25) = 0
16t = 0 t-25 = 0
t = 0 t = 25

v(t) = s'(t) = -32t+400
v(25) = -32(25)+400
= -400

The projectile hits the ground after 25 seconds at a velocity of 400m/s

B s'(t) = v(t) = -32t+400
0 = -32t+400
32t = 400
t = 12.5
s(12.5) = -16(12.5)^2+400(12.5)
= 2500m

The projectile reaches a max altitude of 2500m

C a(t) = v'(t) = -32m/s^2

The acceleration at anytime is 32m/s^2
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  • #2
Huh, I wonder if the original problem had a typo, because the acceleration here is calculated at -32 m/s^2, but on Earth g ≈ -32 ft/s^2. In metric units, g ≈ -9.8 m/s^2.

Besides this curiosity, I see nothing wrong with your work, except for a small typo on the 2nd line of A:
16t^2-400 = 0
should be
16t^2-400t = 0.
  • #3
Well hopefully their answers have a typo as well or I will be way off. Good to know I got the calculations right tho, Thanks for the help

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