A woman throws a ball at a vertical wall d = 6.6 m away. The ball is h = 3.3 m a

Question

A woman throws a ball at a vertical wall d = 6.6 m away. The ball is h = 3.3 m above ground when it leaves the woman's hand with an initial velocity of 14 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)

(a) Where does the ball hit the ground?
_ m (away from the wall)

(b) How long was the ball in the air before it hit the wall?
_ s

(c) Where did the ball hit the wall?
_ m (above the ground)

(d) How long was the ball in the air after it left the wall?
_ s

** This question is really confusing due to the many steps involved, and i could really use some help, thank you!!**

Explanation / Answer

Ans:-

Given the values:
d = 6.6 m
h = 3.3 m
Vo = 14 m/s
= 45°
g = 9.81 m/s^2 (Earth gravity mean)

I solved it in this order. B --> C --> D --> A,

First find the initial and horizontal components of trajectory. Because it's 45° they will be the same, but I'll show it anyway.

Voy = Vo * sin
Voy = (14 m/s) * (sin45)
Voy = (14 m/s) * (0.707)
Voy = 9.898 m/s

Vox = Vo * cos
Vox = (14 m/s) * (cos45)
Vox = (14 m/s) * (0.707)
Vox = 9.898 m/s

Start with (B). Time before impact on the wall. Impact height doesn't matter because horizontal velocity is constant throughout the entire trajectory.

Ti = d / Vox

Ti = (6.6 m) / (9.898 m/s)
Ti = 0.67 s

Move to (C). Height of impact on the wall. Knowing the time it took to hit, and you vertical component, find the impact elevation.

H = h + [ Voy * Ti ] - [ 0.5 * g * Ti^2 ]

H = (3.3 m) + [ (9.898 m/s) * (0.67 s) ] - [ 0.5 * (9.81 m/s^2) * (0.67 s)^2 ]
H = (3.3 m) + [ 6.63 m ] - [ (4.905 m/s^2) * (0.4489 s^2) ]
H = (9.93 m) - [ 2.20 m ]
H = 7.73 m

Move to (D). Time in air after impact with wall. Very simple, knowing height. No horizontal component is necessary, it will fall at the same rate if it's going sideways at 100 m/s or if it's falling at 0 m/s.

Tf = SQRT { [2H] / g }

Tf = SQRT { [ 2 * (7.73 m) ] / (9.81 m/s^2) }
Tf = SQRT { [ 15.46 m ] / (9.81 m/s^2) }
Tf = SQRT { 1.5759 s^2 }
Tf = 1.26 s

Finally get back to (A). Range of impact away from the wall. if you look, this is just basic distance, rate, time that we learned in 6th grade.

R = Tf * Vox

R = (1.26 s) * (9.898 m/s)
R = 12.43 m

(5.83 m behind the woman who threw it)

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