Exactly 3.0s after a projectile is fired into the air from the ground, it is observed to have a velocity v = (8.4 i^ + 4.9 j^)m/s, where the x axis is horizontal and the y axis is positive upward. determine the horizontal range of the projectile. determine its maximum height above the ground. determine the speed of motion just before the projectile strikes the ground. determine the angle of motion just before the projectile strikes the ground.

We are given the equation:

v = (8.4 i^ + 4.9 j^) m/s

Wherein we can break this into the components:

vx = 8.4 m/s

vy = 4.9 m/s

Now solving for the unknowns.

a. Determine the horizontal range of the projectile:

The vx is constant all throughout, therefore the vx at t=0, and vx at t=3 is equal.

x = (8.4 m/s) * 3 s

x = 25.2 m

b. Determine its maximum height above the ground:

We know that vy = 4.9 m/s, therefore we need to find the v0y since this is not constant unlike vx.

vy = v0y – gt

4.9 = v0y – 9.8 * 3

v0y = 34.3 m/s

then find time when height is maximum:

y = v0y * t – 0.5 gt^2

get 1st derivative:

dy / dt = v0y – gt

equate to 0:

v0y – gt = 0

t = 34.3 / 9.8

t = 3.5 s

so the maximum height ymax is:

ymax =  v0y * t – 0.5 gt^2

ymax = 34.3 * 3.5 – 0.5 * 9.8 * (3.5)^2

ymax = 60.025 m

c. Determine the speed of motion just before the projectile strikes the ground.

The final velocity is equally similar to the initial velocity

v^2 = v0x^2 + v0y^2

v^2 = 8.4^2 + 34.3^2

v = 35.31 m/s

d. Determine the angle of motion just before the projectile strikes the ground.

We use the tan function to determine the angle

tan θ = vy / vx

θ = tan^-1 (34.3 / 8.4)

θ = 76.24°

Summary of answers:

a. x = range = 25.2 m

b. ymax = max height = 60.025 m

c. v = 35.31 m/s

d. θ = 76.24°