Question

A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m from where it was launched.

-Find the hang time for the projectile.
-Find the initial speed of a projectile.
-What are the x and y components of the projectile’s velocity the moment before it strikes the ground?
-At what speed will the projectile strike the ground?

Answer 1

a)t = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Step-by-step explanation:

According to the problem statement

V₀ = V₀ₓ and V₀y = 0

And at the end of the movement t = ? the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9 ⇒ t² = 2,04 s

t = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then 15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf = 0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s