PHYS 130 Prof. Finn

HW Solutions: Ch 4

P4.10 To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 300m/s at 55.00 above the horizontal. It explodes on the mountainside 42.0 s after firing. What are the x andy coordinates of the shell where it explodes, relative to its firing point?

Given:

vi = 300 m/sθ = 55.00

t = 42.0 s

Find:

xf

yf

Set origin (0,0) to where the shell is launched. Then:xi 0xf ?vxi vicosθax 0

yi 0yf ?vyi visinθay −9.8 m/s

Solve x motion:xf = xi + vxit + 1

2axt2

xf = 0 + vicosθt + 0

xf = 0 + (300 m/s) cos(550) (42.0 s)

xf = 7.23 × 103 m

Solve y motion:yf = yi + vyit + 1

2ayt2

yf = 0 + (300 m/s)sin(550)(42.0 s) + 12 (−9.8 m/s2)(42.0 s)2

yf = 1.68 × 103 m

P4.12 In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender ismomentarily distracted and does not see the mug, which slides off the counter and strikes the floor a distanced from the base of the counter. The height of the counter is h. (a) With what velocity did the mug leavethe counter, and (b) what was the direction of the mug’s velocity just before it hit the floor?

Set origin of x and y axes to the point where the counter meets the floor. Then:Given:

xf = dyi = h

1

Find:

vxi

direction of vf

xi 0xf dvxi ?ax 0

yi hyf 0vyi 0ay −g

where g = 9.8 m/s2.

(a) Time-of-flight is determined by y-motion, so first solve for time using the y equation for position as afunction of time:

yf = yi + vyit + 12ayt2

0 = h + 0 − 12gt2

−h = − 12gt2

2hg = t2

t =√

2hg

Now that we have an expression for time, we can input that into the x equation for position as a function oftime and solve for vxi.

xf = xi + vxit + 12axt2

d = 0 + vxi(√

2hg ) + 0

vxi = d√

2hg

(b) To find direction of final velocity, first find components, vxf and vyf .

The acceleration in the x-direction is zero, so:vxf = vxi + axt

vxf = vxi = d√

2hg

We have acceleration in the y-direction, so:vyf = vyi + ayt

vyf = 0 − g√

2hg

vyf = −g√

2hg

Now we can find the angle of the final velocity:tanθ =

vyf

vyi

θ = tan−1(vyf

vyi)

2

θ = tan−1(−g

√

2hg

d√2hg

)

θ = tan−1(−g 2h

g

d )

θ = tan−1(−2hd )

P4.19 In football, a place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal,and half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leavesthe ground with a speed of 20.0 m/s at an angle of 53.00 to the horizontal. (a) By how much does the ballclear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or whilefalling?

Set origin of x and y axes to the point where the ball is kicked. Then:

Given:

xf = 36.0vi = 20.0 m/sθ = 53.00

Find:

yf

where does ball reach ymax?

xi 0xf 36.0 mvxi vicosθax 0

yi 0yf ?vyi visinθay −g

where g = 9.8 m/s2.

(a) Find the time it takes to reach xf = 36.0 m by solving the x-equation of motion:xf = xi + vxit + 1

2axt2

xf = 0 + vi cos(θ)t + 0

t =xf

vi cos(θ)

t = 36.0 m20.0 m/s cos(53.00)

t = 2.99 s

Now find the ball’s y-height at this time.

yf = yi + vyit + 12ayt2

yf = 0 + vi sin(θ)t − 12gt2

yf = 0 + 20.0 m/s sin(53.00)(2.99 s) − 12 (9.8 m/s2)(2.99 s)2

yf = 0 + 20.0 m/s sin(53.00)(2.99 s) − 12 (9.8 m/s2)(2.99 s)2

3

yf = 3.95 m

Therefore, the ball clears the post by 3.95 m − 3.05 m = 0.90 m .

(b) Now find time to reach max height and compare this with time from part (a). If the football reaches itsmaximum height before 2.99 s, then the ball will be on its way down. If the football reaches its maximumheight after 2.99 s, then the ball will still be on its way up.

At maximum height, the y-component of the ball’s velocity is zero. So, use the y equation for velocity as afunction of time to find the time corresponding to maximum height.

vyf = vyi + ayt

t =vf−vi

ay

t = 0−vi sin(θ)−g

t = −(20.0 m/s) sin(53.00)−9.8 m/s2

t = 1.63 s

So ball clears the goal the post after maximum height when it is on its way down.

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