Projectile Problems with Solutions 5

Problem#1
As some molten metal splashes, one droplet flies off to the east with initial velocity v_i at angle θ_i above the horizontal, and another droplet to the west with the same speed at the same angle above the horizontal, as in Figure 1. In terms of v_i and θ_i , find the distance between them as a function of time.

Fig.1

Answer:

At any time t, the two drops have identical y-coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore,

d = 2|x(t)| = 2v_xit

with v_xi = v_i cos θ_i, then

d = 2v_it cos θ_i

Problem#2

Fig,2

A projectile is fired up an incline (incline angle .) with an initial speed vi at an angle 'i with respect to the horizontal (θ_i > φ), as shown in Figure 2. (a) Show that the projectile travels a distance d up the incline, where

d = [2v_i² cos θ_i sin (θ_i – φ)]/[g cos² φ]

₍b) For what value of θ_i is d a maximum, and what is that maximum value?


Answer:
(a) we use

y_f = x_f tan θ_i – ½ [gx_f²/(2v_i² cos² θ_i)]

Setting x_f = d cos φ , and y_f = d sinφ , we have

d sin φ = (d cos φ) tan θ_i – ½ [g/(2v_i² cos² φ)][d cos φ]²

Solving for d yields,

sin φ = (cos φ) tan θ_i – ½ [g/(2v_i² cos² θ_i)][d cos² φ]

(cos φ) tan θ_i – sin φ = ½ [g/(2v_i² cos² θ_i)][d cos² φ]

cos φ sin θ_i – sin φ cos θ_i  = ½ [g/(2v_i² cos² θ_i)][d cos² φ] cos θ_i

sin (θ_i – φ) = ½ [g/(2v_i² cos θ_i)][d cos² φ]

2v_i² cos θ_i sin (θ_i – φ) = [gd cos² φ]

d = [2v_i² cos θ_i sin (θ_i – φ)]/[g cos² φ]

(b)  the maximum value θ_i when

dd/dθ_i = 0

Let us note that for given v_i, φ, g (and kept constants) the only factor in the expression for d that depends on θ is sin(θ − φ) cos θ. This can be rewritten as

sin(θ_i − φ) cos θ_i = ½ {sin(θ_i − φ + θ_i) + sin(θ_i − φ – θ_i)}

= ½ {sin(2θ − φ) − sin(φ)}

Then

dd/dθ_i = cos (2θ_i − φ) = 0

cos (2θ_i − φ) = cos π/2

2θ_i – φ = π/2

θ_i = π/4 + φ/2

the maximum value of d is found to be

d = [2v_i² cos {π/4 + φ/2} sin (π/4 + φ/2 – φ)]/[g cos² φ]

d_max = [v_i² (1 – sin φ)]/[g cos² φ]


Problem#3
Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, 21.0 m high, located 130 m from home plate. The ball is hit at an angle of 35.0° to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time at which the ball reaches the cheap seats, and (c) the velocity components and the speed of the ball when it passes over the top row. Assume the ball is hit at a height of 1.00 m above the ground.

Answer:
Refer to the sketch, Fig.3:

Fig.3

(b) the time at which the ball reaches the cheap seats,

we use
∆x = v_xit

130 m = (v_i cos 35.0⁰)t

v_i = 130/(t cos 35.0⁰)        (*)

and

y = y_i + v_yit + ½ a_yt²

21 m = 1.0 m + (v_i sin 35.0⁰)t + ½ (–9.80 m/s²)t²

substitution (*) then

20 m = (130/t cos 35.0⁰)(sin 35.0⁰)t + ½ (–9.80 m/s²)t²

20 m = 130 tan 35.0⁰ – 4.90t²

–71.0269 = –4.90t²

t² = 14.495

t = 3.81 s

(a) then the initial speed of the ball is

v_i = 130/(t cos 35.0⁰) = 130 m/(3.81 s cos 35.0⁰) = 41.7 m/s

(c) the velocity components and the speed of the ball when it passes over the top row

At t = 3.81 s:
v_xf = v_xi = v_i cos 35.0⁰ = (41.7 m/s) cos 35.0⁰ = 34.1 m/s

and

v_yf = v_yi – gt = v_i sin 35.0⁰ – gt

v_yf = 41.7 m/s sin 35.0⁰ – (9.80 m/s²)(3.81 s) = –13.4 m/s

then

v_f = √[v_xf² + v_yf²]

v_f = √[(34.1 m/s)² + (–13.4 m/s)²] = 36.7 m/s

Problem#4
A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure 4. If he shoots the ball at a 40.0° angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.

Fig.4

Answer:

Given: if we take the floor reference point then, y_i = 2.00 m, y_f = 3.05 m, x = 10.0 m and θ = 40.0⁰.

the velocity component on the x-axis and the velocity on the y-axis are

v_xi = v_i cos 40.0⁰ = v_i(0.766) and v_yi = v_i sin 40.0⁰ = v_i(0.643)

the position of the ball on the x-axis is

x_f = x_i + v_xit

10.0 m = 0 + v_i(0.766)t

t = 10.0/(0.766v_i)                              (*)

the position of the ball on the y-axis is

y_f = y_i + v_yit + ½ a_yt²

3.05 m = 2.00 m + v_i(0.643)t + ½ (–9.80 m/s²)t²

1.05 = v_i(0.643)t –4.90t²

substitution t of (*),

1.05 = v_i(0.643)[10.0/(0.766v_i)] –4.90[10.0/(0.766v_i)]²

1.05 = 8.394 – 4.90[10.0/(0.766v_i)]²

1.499 = [10.0/(0.766v_i)]²

1.224 = 10.0/(0.766v_i)

From this, v_i = 10.7 m/s   

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