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°
