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 34° with respect to the horizontal and with an initial speed of 25 m/s. Air resistance is negligible in this situation.

1-What is the maximum height that the projectile reaches?

2-What is the speed of the projectile at its maximum height?

3-The same projectile was then fired in the same way toward a wall that is a horizontal distance 49.6 m from where the projectile was fired.
What was the height of the projectile when it hit the wall?

4-What was the velocity of the projectile when it hit the wall?
Call down the positive y direction, and toward the wall the positive x direction. ...i+..j

If the wall were slightly farther from where the projectile was launched (the projectile still hits it), how would that effect the following? The maximum height of the projectile. The magnitude of the y component of the velocity of the projectile when it hits the wall. The speed of the projectile when it hits the wall. The distance from the ground to where the projectile hit the wall. The magnitude of the x component of the velocity of the projectile when it hits the wall.

Explanation / Answer

(A) In vertical,

initial vertical velocity, v0y = 25 sin34 = 13.98 m/s

ay = - g = - 9.8 m/s^2

at highest point, vertical velcoity will be zero.

Applying vf^2 - vi^2 = 2 a d

0^2 - 13.98^2 = 2 (-9.8) (Hmax)

Hmax = 9.97 m

(B) at highest point, vertical component will be zero so there will be only horizontal component.( that is constant throughout)

v = v0x = 25 cos34 = 20.725 m/s

(c) X = v0x t

t = 49.6 / (20.725) = 2.393 sec

y = v0y t + ay t^2 /2

H = (13.98 x 2.393) - (9.8 x 2.393^2 / 2)

H = 5.40 m

(D ) vx = 20.725 m/s

vy = 13.98 - (9.8 x 2.393) = -9.47 m/s

v = 20.72 i - 9.47 j

(e) speed = magnitude of v = sqrt(vx^2 + vy^2 ) = 22.8 m/s

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