What is the maximum height reached while in flight ?

Question

A cannon fires a projectile with a velocity of 300.1m/s at an angle of 32 degrees above the horizontal. What is the maximum height that the projectile reaches while in flight ?

Answer 1

The first step is to determine the vertical speed of the projectile.

V_h = 300.1 x sin 32⁰ = 159 m/s

For objects freely falling with an initial velocity use:

V_f² = V_i²+ 2ay V_i is the initial vertical speed (Up is positive) ; a= -9.81m/s²

V_f is the final speed. For this problem at max height vertical speed is 0. Y is vertical height.

0² = (159)² + 2 (-9.81)y Solve for y. y= 1288.1 m Hope this helps.

Answer 2

This a classic high school physics problem: a projectile in free fall with some initial velocity at some angle.(As is standard for text physics, I will be bolding vectors. For example, the acceleration vector is a but the magnitude of the acceleration is a)

First, let's find the x and y components of the initial velocity. Recall from trigonometry that x-component is the initial velocity multiplied by the cosine of the angle and the y-component is the same, but multiplied by the sine of the angle. (Note that this is true only for the angle made with respect to the horizontal. If the angle was made with respect to the vertical, we would swap the trig. functions.)

v_i_x = initial x-component of the velocity = v_i * cos(theta) = 300.1 * cos(32) = +254.5 m/s

v_i_y = initial y-component of the velocity = v_i * sin(theta) = 300.1 * sin(32) = +159.0 m/s

Next, I had to make some assumptions since the question is vague. Let's assume that we neglect air resistance so the acceleration of the projectile is constant. Next, understand that the max height I calculate here is the max change in the y-component of the position; that is, even if the cannon/projectile did not start from the ground, this value would still be the max height, just not measured from the ground. And I use the value of 9.81 m/s^2 for the value of g.

Finally, we can use our kinematics equations (we have the acceleration, and initial velocity) and the fact that at the max height for an object in free fall the y-component of the velocity is 0.

(v_f_y)^2 = (v_i_y)^2 +2ad

0 = (v_i_y)^2 +2ad

(v_i_y)^2 = -2ad

d = (v_i_y)^2 / -2a = (v_i * sin(theta))^2 / (-2 * -g) = (159.0)^2 / (2 * 9.81) = 1288.5 m

Therefore, our max height is 1288.5 m above where we started.

Thank you for your question and I hope this helps!

-MJM