Question

1. A projectile is launched horizontally from a height above level ground of 3 m. When it hits the ground, it's horizonta displacement from the launch point is 7 m. a. b. How much time does it spend in flight? With what speed was it launched? Suppose the projectile were now launched from the same point at an angle of 45 above the horizontal. a. b. How much time does it spend in the air? What is its horizontal displacement from the launch point when it hits the ground?

Answer 1

Ans:

1)

a) di = 3 m

     df = 0 m

      g = -9.8 m/s²

      vi = 0

So,

df = di + vi(t) + ½(gt²)

t = 2.21 s

b) vf = d/t = 7/2.21 = 9.48 m/s

2)

            a) when projectile is in the air, it rises and then falls to a final possition 3 m lower than its starting altitude. We can find the time by using

y = y₀ + v_0yt – 1/2gt²

where y₀ = 0

y = -3 m

Initial vertical velocity

V0y = v0sinϴ = 6.68 m/s

Substituting known values yields

-3 = 6.68 t – 4.9 t2

Solving this eq.

We get

T = 1.7193 s

            b) Horizontal displacement = v*t = 11.4886 m