The basketball passed through the hoop even though it barely cleared the hands of the player...

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The basketball passed through the hoop even though it barely cleared the hands of the player...

The basketball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Suppose that s = 3 ft . Neglect the size of the ball.

Part A

Determine the magnitude vA of its initial velocity.

Express your answer to three significant figures and include the appropriate units.

Part B

Determine the height h of the ball when it passes over player B.

Express your answer to three significant figures and include the appropriate units. no title provided

12.95

engineering Mechanical-Engineering

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Answer 1

Answer #1

Concepts and reason

The body when projected in a two dimensional plane, it will have velocity in both x and y directions. Gravitational acceleration is in the y direction and in the x direction acceleration is zero.

The maximum height that the object will reach when thrown is called the maximum height of the projectile.

The time for which the projectile remains in the air when thrown is called the time of flight.

Fundamentals

The figure (1) shows the two dimensional projectile motion.

|uy
Hmax
A
ux
R
Figure 1

The components of acceleration along x and y directions respectively are,

a 0
a,=-g

The components of velocity along x and y directions respectively are,

и, 3исos0
и, 3usin 0
и

Here, the velocity of projection is u and the angle of projection is .

The equation of motion along x-direction is,

x=u,t

Here, the time of flight is t.

The equation of motion along y-direction is,

у-ит-
2

Here, the acceleration due to gravity is g.

The motion of a projectile is shown in Figure (2).

(Position of the
attempt of the Player)
В
C (Position of Basket)
300
А
h
10 ft
7 ft
5 ft
25 ft
Figure 2

Consider the ball’s motion along horizontal direction in figure (2).

The equation for the ball to reach B from A from figure (2) is written as,

|(V,cos 30e)4 25

Here, time taken by the ball from A to reach B is .

0.866V 25
25
V0866
0.866

V28.87 …… (1)

The equation for the ball to reach C from A from figure (1) is written as,

(V, cos 30), (25+3) ft

Here, time taken by the ball from A to reach C is .

0.866Vt 28
28
V2 0.866

V=32.33 …… (2)

Divide equation (2) by (1),

V 32.33
VA 28.87

2=1.12
4 …… (3)

Consider the ball’s motion along vertical direction in figure (2).

The equation for the ball to reach C from A from figure (2) is written as,

1
(V, sin 30°), +7-8 =10
gt = 10
2

Here, acceleration due to gravity is g.

Substitute 32.2 ft/s for g

0.5V2+7-0.5x32.2 10
0.5V2-16.110-7
16.12-0.5V+3 0

…… (4)

The equation for the ball to reach B from A from figure (2) is written as,

|(V, sin 30°)4 +7-gf h

Substitute, 32.2 ft/s for g

0.5V+7-0.5x32.2xt? =h
V14-32.2 2h

32.2-V14-2h …… (5)

Substitute 28.87 for from equation (1) in equation (5)

32.22-28.87 14-2h

…… (6)

Substitute 32.33 for from equation (2) in equation (4)

2-0.031x32.33+0.186=0
2-0.81623 0
0.81623
t2 =0.903 s

Write the equation (3),

2=1.12
4

Substitute 0.942 s for

0.9031.12
4 =0.806 s

(a)

Calculate the initial velocity of the ball from equation (1).

Substitute 0.792 s for

V2(0.806 28.87
35.81 ft/s

(b)

Calculate the height of the ball when it passes over player B from equation (6).

Substitute 0.806 s for .

32.2x (0.806s)42.87-2h
2h 21.95
h 10.9 ft

Ans: Part a

The initial velocity of the ball (V) is 35.81 ft/s .

Part b

The height of the ball (h) when it passes over player B is 10.9 ft .

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