Max velocity and max height!

Question

A object is launched into the air vertically from ground level with initial velocity 250 m/s.

What is the object's maximum velocity?    m/s

What is the object's maximum height?    m

Answer 1

Since the direction of motion of the object is the same as the direction of its acceleration due to gravity (vertical), we can approach this as a one-dimensional problem in the vertical (y) direction. We can (arbitrarily) choose y to be zero at ground level and positive pointing in the upward direction away from the ground.

The acceleration, a, of the object in this coordinate system would then be the acceleration due to gravity g = -9.8 m/s^2. This acceleration is equal to the rate of change of the velocity, dv/dt. So we now have the equation dv/dt = g. We can integrate this equation with respect to time (t) because g is a constant. This gives us an equation for velocity, v(t) = g*t + A, where A is the constant of integration. We know that the initial velocity is 250 m/s, and we can plug this into our velocity equation to get v(t=0) = 250 = g*0 + A, which simplifies to show us that 250 = A. So v(t) = g*t + 250.

Now, the velocity is the rate of change of the object's position, dy/dt. We can integrate the equation dy/dt = g*t + 250 to find that y(t) = 1/2*g*t^2 + 250*t + B, where B is our constant of integration. We can solve for B since we know that the object starts on the ground at y(t=0) = 0 = 1/2*g*0 + 250*0 + B. This then simplifies to give us B = 0.

We now have equations for position and velocity:

y(t) = 1/2*g*t^2 + 250*t

v(t) = g*t + 250

Now let's solve for the maximum height first. The maximum height occurs when the velocity goes from being positive (upward motion) to being negative (downward motion). This is the same as saying that it occurs when the velocity equals zero, so we can set v(t_m) = 0 = g*t_m +250, and then solve for the time at which the height is greatest, t_m. This gives us t_m = -250/g. We can then plug this value into the equation for y(t) to find the height of the object at that time. This would look like y(t_m) = 1/2*1/g*250^2 - 250*250/g. (Don't forget that g is negative when you plug in the numbers!)

I'm going to assume that the question is actually looking for the maximum speed of the object (absolute value of velocity) since the maximum velocity would simply be 250 based on our equation for v(t) (since g is negative, the velocity only decreases from its initial value). This is actually the correct answer since the symmetry of the parabola defined by the equation y(t) ensures that the speed of the object when it hits the ground is the same as its speed when it is first launched, but we can go through the exercise to prove that.

The speed of the object must be at its maximum either at the moment it is launched (since it will immediately begin slowing down as soon as it starts traveling upwards) or at the moment it hits the ground after falling (since it accelerates downward the entire time it falls from its peak height). So we need to know how much time it takes for the object to hit the ground. We can find this by setting y(t_f) = 0 = 1/2*g*t_f^2 + 250*t_f. We can solve this by factoring out t_f on the right hand side to give 0 = t_f ( 1/2*g*t_f + 250 ). One solution to this equation is of course t_f = 0 since the object starts on the ground initially. The other solution is t_f = -500/g. Notice that we could have gotten this value by recognizing that the time to fall from the max height is the same as the time to reach the max height, so t_f = 2*t_m because of the symmetry of the parabola. Now we can plug t_f into v(t) to find that v(t_f) = -500 + 250 = -250. The speed is just the absolute value, so we have found that the initial and final speeds of the object are both the maximum value of 250 m/s.

Answer 2

Let us set our coordinate plane on the ground with going up as positive.

Initial velocity means that at time t = 0, the velocity is 250 m/s (or that the point 90, 250) is a point on the velocity function graph.

Also, since we are initially launching from ground level and ground level has a height of 0, we can say s(0) = 0 or the point (0, 0) is a point on the position function graph.

The derivative of position (s) is velocity so:

ds/dt = v(t) therefore

ds = v(t) dt or

∫ds = ∫v(t) dt or

s(t) = ∫v(t) dt

We also know that:

dv/dt = a(t) or

dv = a(t) dt or

∫dv = ∫a(t) dt or

v(t) = ∫a(t) dt

On earth the acceleration due to gravity is -9.8 m/s² (9.8 m/s2 pointing downward means it is negative), therefore:

v(t) = ∫(-9.8) dt

v(t) = -9.8t + C

Using the velocity graph point (0, 250) we can solve for C:

v(0) = 250 = -9.8(0) + C so C = 250

This makes our velocity function:

v(t) = -9.8t + 250

We can now plug this into

s(t) = ∫v(t) dt to make

s(t) = ∫(-9.8t + 250) dt and integrate to get:

s(t) = -4.9t² + 250t + C

and using the point (0, 0) we can solve for C:

0 = -4.9(0)² + 250(0) + C or C = 0 meaning our position function is:

s(t) = -4.9t² + 250t + 0 or

s(t) = -4.9t² + 250t

Now, to answer the questions.

To find the max velocity, use v(t) = -9.8t + 250

It's easy to see that the max velocity occurs when t = 0, every value larger than zero means you subtracting something away from the 250.

At t = 0, the velocity is 250 m/s

To find the max height, use the position function:

s(t) = -4.9t² + 250t

and take the derivative and set it equal to zero to maximize it:

s'(t) = -9.8t + 250

-9.8t + 250 = 0

250 = 9.8t

t = 1250/49

Plug this into the position equation to get the height:

s(1250/49) = approx 3190 meters