The distance s (in m) above the ground for a projectile fired... Please see description for the rest

Question

The distance s (in m) above the ground for a projectile fired vertically upward with a velocity of 40 m/s as a function of time t (in s) is given by s=40t-4.9t^2. Find the answers for these questions. find t for v=0 , find v for t=4 , find v for t=5, what conclusions can you draw

Can I please get the work shown and broken down as I am sort of lost on this. Thank you

Answer 1

Well, if it is fired absolutely vertically upwards, this equation won't work.

So it is fired upwards at a certain angle, thus the equation is s = 40t - 4.9t²

Graphing it on a calculator, s = 81.6 m at it highest elevation which is when t = 4.08 seconds.

It reaches the ground level when t = 8.16 seconds.

It's arc is an upside down parabola.

If you use calculus and find the first derivative of the equation, it gives the maximum height which

calculates out as : 0 = 40 - 9.8t or t also equals 4.08 seconds.

And setting the formula = 0 yields two answers when it is at ground level, t=0 and t = 8.16 seconds

Answer 2

The first derivative with respect to time of the position function S(t) is the velocity function V(t).

S(t)=40t - 4.9 * t²

S'(t)=V(t)= 40 - 2* 4.9 * t

V(t)= 40 - 9.8 * t.........................Eq. 1

Find "t" for v=0m/s

Plugging in V=0 into Eq. 1 and solving for t:

0= 40 - 9.8 * t

9.8 * t= 40

t= 4.082 seconds

Find "v" for t= 4s

Plugging t=4 into Eq. 1 and solving for v:

v= 40 - 9.8 * 4

v= 0.8m/s

Find "v" for t=5s

Plugging t=5 into Eq. 1 and solving for v:

v= 40 - 9.8 * 5

v= -9m/s

What conclusions can be drawn?

-The ball reached its maximum height at t=4.082s. Before this time, the balls velocity was upwards, after this time, the velocity was downwards.