A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no obstacles in its way. It is fired at an angle 33° with respect to the horizontal and with an initial speed of 49 m/s. Air resistance is negligible in this situation.

A) What is the horizontal distance that the projectile has traveled when it is at its maximum height?

B) The same projectile was then fired in the same way toward a wall that is a horizontal distance 55.8 m from where the projectile was fired.

What was the height of the projectile when it hit the wall?

C) Find the velocity of the projectile when it hit the wall.

D) What was the speed of the projectile when it hit the wall?

Question

Emery Gillespie

Physics

9 March, 05:00

A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no obstacles in its way. It is fired at an angle 33° with respect to the horizontal and with an initial speed of 49 m/s. Air resistance is negligible in this situation.

A) What is the horizontal distance that the projectile has traveled when it is at its maximum height?

B) The same projectile was then fired in the same way toward a wall that is a horizontal distance 55.8 m from where the projectile was fired.

What was the height of the projectile when it hit the wall?

C) Find the velocity of the projectile when it hit the wall.

D) What was the speed of the projectile when it hit the wall?

+4

Answers (1)

Know the Answer?

Rhett Lewis · Answer 1

(A) 111.77m

(B) 9.07m

(C) 55.88m/s

(D) 20.54m/s

Explanation:

step 1 " we have to calculate the time it took the projectile to get to its maximum height

(a) t = usinθ/g

= 49sin 33/9.81

49*0.5446/9.81

=2.72s

the horizontal distance = ucosθ*t, because the projectile horizontal motion is unaffected by the force of graavity

= 49cos33 * 2.72

=111.77m

(B) with the same projectile fired the same way, the horizontal distance = 55.8m

55.8 = ucosθ*t

55.8 = 49cos33 * t

t = 55.8/49cos33

t = 1.36s

height of the projectile = 1/2 * g*t²

=1/2 * 9.81*1.36²

= 9.07m

(c) Velocity of the projectile when it hits the wall

V₀ = ucosθ*t

49cos 33 * 1.36

=55.88 m/s

(D) speed = distance / time

distance / 2*t; total time of flight

= 55.88 / 2.72

=20.54m/s