A projectile is fired with an initial speed of 30 m/s at an angle of 60° above the horizontal. The object hits the ground 7.5 seconds later.

Question

1)How much higher or lower is the launch point relative to the point where the projectile hits the ground?

2)To what maximum height above the launch point does the projectile rise?

3)What are the magnitude and direction of the projectile's velocity at the instant it hits the ground?

Answer 1

First break the initial velocity vector into its components v_i-x and v_i-y like this:

where v_i-x = 30cos(60°) and v_i-y = 30sin(60°)

Now, consider what happens in the vertical direction. The initial vertical velocity is again v_i-y = 30sin(60°) = 25.9808 m/s

An object traveling upwards with an initial velocity will slow to a stop due to the force of gravity. The acceleration due to gravity is -9.81 m/s and we can use the kinematic equation v_f = v_i + at to calculate the time it takes for the object to slow to a stop (top of its path):

v_f = v_i + at

0 = 25.9808 + (-9.81)t

9.81t = 25.9808

t = 25.9808/9.81 = 2.6484 seconds

The distance it travels in those 2.6484 seconds can be calculated using the kinematic equation x = x_i + v_it + 1/2at² where x_i = 0, v_i = 25.9808, t = 2.6484, a = -9,81:

x = x_i + v_it + 1/2at²

x = 0 + (25.9808)(2.6484) + (1/2)(-9.81)(2.6484²) = 34.4037 m

Now the object begins to fall. The time rising was 2.6484 seconds and the total time was given as 7.5 seconds so the time falling is 7.5 - 2.6484 = 4.8516 seconds

Using the kinematic equation of motion x = x_i + v_it + 1/2at² where x_i = 0, v_i = 0 (top of the flight the object is instantaneously motionless in the y-direction), t = 4.8516 s, and a = -9.81 we get:

x = x_i + v_it + 1/2at²

x = 0 + (0)(4.8516) + (1/2)(-9.81)(4.8516²)

x = -115.4542 m

Since the object rose 34.4037 m and fell 115.4542 m, the object landed 34.4037 - 115.4542 = -81.0505 m from it's initial y-value position in other words it landed 81 m below its launch position.

I'll let you choose whether you want to answer with 1 sig fig or 2. The initial velocity is given as 1 sig fig so in reality, you should answer as 80 m below the launch position.

To find the magnitude and direction of the velocity vector when the object hits, use the v_i-x (because irt stays constant) and calculate the v_f-y using v_f = v_i + at² where v_i = 0 (velocity in the y-direction ot the top of the object's flight), a = -9.81, and t = 4.8516 seconds.

Once you find v_f-y combine it with vf-x using the Pythagorean Theorem a to get the magnitude. Use tan^-1(v_f-y/v_f-x) to get the angle