A projectile is fired from the ground (you can assume the initial height is the

Question

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.
You will enter your answer using the unit vector notation where the value in front of the i is the x component and the value in front of the j is the y component.
Call up the positive y direction, and toward the wall the positive x direction.

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

Explanation / Answer

A)

vi= 49m/s

vix=vi*cos33 = 49cos33 = 41.1m/s

viy=vi*sin33 = 49sin33 = 26.7m/s

At max height viy= 0m/s

Let us find time ‘t’ for max height

vfy= viy+a*t = viy - gt

0 = 26.7-9.8*t         => t =4.2s

Horizontal distance at max height, d = vix*t = 41.1*2.72 = 111.8 m

B)

Here let us find ‘t’ first

d = vix*t

55.8 = 41.1*t    =>   t = 1.36 s

Height at t= 1.36 s

H= viy*t +1/2at^2 = 26.7*1.36 – 1/2*9.8*1.36^2 = 27.25m

C)

vfx=vix = 41.1m/s

vfy= viy+a*t = viy - gt = 26.7 - 9.8*1.36 = 13.37 m/s

vf = vfx + vfy = 41.1i + 13.37j

D)

vf = sqrt(vfx^2 + vfy^2) = sqrt(41.1^2 +13.37^2) = 43.22 m/s

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