Determine the acceleration due to gravity on this planet

[darbeecakes]

Homework Statement

A projectile is launched over level ground on an unnamed planet. The equation of the projectile's vertical position is y(t)=-6t²+48t-90.
a. Determine the acceleration due to gravity on this planet
b. Determine (y_max, t_ymax), the coordinates of the maximum height of the projectile in the t-y plane

Homework Equations

r_f=r_i+v_it+1/2gt²

The Attempt at a Solution

I know r_i=0 because that's the origin
Since I'm not given a time or velocity I'm stuck on how to solve for anything

 

[ehild]

Homework Statement

A projectile is launched over level ground on an unnamed planet. The equation of the projectile's vertical position is y(t)=-6t²+48t-90.

Homework Equations

r_f=r_i+v_it+1/2gt²

The Attempt at a Solution

I know r_i=0 because that's the origin
Since I'm not given a time or velocity I'm stuck on how to solve for anything

Write y and y_i in your equation instead of r and r_i. The equation given and your one should be equivalent, that is they have to give the same y at every time t. What does it mean for y_i, v_i and g?

ehild

 

[darbeecakes]

Thank you. y_i is the initial y position, v_i is the initial velocity, and g is gravity due to acceleration.

 

[ehild]

Thank you. y_i is the initial y position, v_i is the initial velocity, and g is gravity due to acceleration.

Compare it with the original equation
y(t)=-6t²+48t-90

What is the value of the initial position, initial velocity and acceleration?

ehild

 

[darbeecakes]

So the initial position is -90, initial velocity is 48t and acceleration is -6t²?

 

[ehild]

t is the running time. Anything that contains t changes with time. The initial velocity is the velocity at t=0. ehild

 

[darbeecakes]

So do I leave the acceleration value as -6t² or do I solve it further by substituting a value for t? And how do I find the final t value?

 

[ehild]

You have the general equation for the vertical position of a projectile

y(t)=y_i+v_it -g/2 t^2

(assuming that the positive y-axis points upward and vi is the initial upward velocity).

On that planet, you launch a projectile upward from a valley and you get the vertical position as function of time:

y(t)=-90 +48t -6t².

These equations should be equivalent, the positions y(t) given by both equations are the same at any time. You certainly have learned that the coefficients of two polynomials have to agree pairwise in order they are equivalent. The coefficients of t^2 in both equations have to be the same: so -g/2=-6. The coefficient of t should agree, too, so 48=v_i.

ehild

 

[darbeecakes]

How do I find the coordinates for the maximum height of the projectile?

 

[ehild]

How can you find the maximum of a function? ehild

 

[darbeecakes]

I know the equation for the maximum height is the same as the first one i gave with ymax substituted in for rfinal and yinitial substituted in for rinitial, but I'm not sure how to find coordinate numbers when two variables are unknown.

 

[ehild]

You should calculate the maximum height and the time the projectile reaches it. If you studied calculus you know that the maximum of y is at that value of t where the derivative of y with respect to t is zero: dy/dt=0.
The other approach: you know that the vertical velocity of a projectile is v=v_i-gt. You can find the rise time from the condition v=0.

ehild

 

[darbeecakes]

I ended up getting t=4 and y=6.
I think it's correct.
Thank you for the help.