Problem 18

A students stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0-m above a flat horizontal beach. How long after being released does the stone strike the beach below the cliff? With what speed and angle of impact does it land?

SOLUTION:

KNOWNS / UNKNOWN
yo = 0 / t
y = -50.0 m
qo = 0
vo = 18.0 m/s

y = yo + (vo sinqo )t – ½ gt2

-50.0 m = 0 + 0 – ½ (9.80 m/s2)t2

t2 = 2(50.0 m)/( 9.80 m/s2)

t = 3.19 s

KNOWNS / UNKNOWN
yo = 0 / v
y = -50.0 m
qo = 0
vo = 18.0 m/s
t = 3.19 s

To find the magnitude and direction of the velocity means finding both x and y components of the velocity vector:

vx = (vo cos qo) = 18.0 m/s

vy = (vo sin qo) – gt = 0 – (9.80 m/s2)(3.19 s) = -31.26 m/s

|v| = [(18.0)2 + (-31.26)2]1/2 = 36.1 m/s

qo = tan-1 (-31.26/18.0) = -60.10

Problem 20

A place kicker must kick a football from a point 36.0 m from the goal. The ball must clear the crossbar which is 3.05 m above the ground. When he kicks the ball, it leaves with a speed of 20.0 m/s at an angle of 53.00 to the horizontal.

(a) By how much does the ball clear or fall short of the crossbar?

SOLUTION:

KNOWNS / UNKNOWNS
x = 36.0 m / y
qo = 53.00
vo = 20.0 m/s
yo = 0

y = yo + (vo sinqo )t – ½ gt2

But we do not know what t is so we need to find t first:

x = (vo cos qo)t

36.0 m = (20.0 m/s)(cos 53.0)t

t = 2.99 s

y = yo + (vo sinqo )t – ½ gt2

y = 0 + (20.0 m/s)(sin53.0)(2.99) – ½ (9.80 m/s2)(2.99 s)2

y = 47.76 m – 43.81 m = 3.95 m

The ball does clear the bar by 0.90 m.

b) Does the ball approach the cross bar while still rising or falling?

SOLUTION:

To do this, we need to find the time it takes to reach maximum height. If the time it takes to reach maximum height is less than 2.99 s then the ball was rising, otherwise, the ball is falling.

At maximum height, vy = 0, therefore:

vy = (vo sin qo) – gt

0 = (20.0 m/s)(sin 53.0) – (9.80 m/s2)t

t = 1.63 s therefore FALLING.

Problem 23

A projectile is launched with an initial speed of 60 m/s at an angle of 300 above the horizontal. The projectile lands on a hillside 4.0-s later. Neglect air friction.

(a) What is the projectile’s velocity at the highest point of its trajectory?

SOLUTION:

To find the velocity of a projectile, we must find the velocity in the x and y directions, then find the magnitude and direction of the velocity. This particular question wants the velocity at maximum height and we know that at maximum height, the velocity in the y-direction is equal to ZERO.

KNOWNS / UNKNOWNS
t = 4.0 s / v
qo = 300
vo = 60 m/s
vy = 0

vy = 0

vx = (vo cos qo) = (60 m/s)(cos 300) = 52 m/s

|v| = 52 m/s at 0 degrees to the horizontal.

(b) What is the straight-line distance from where the projectile was launched to where it hits?

KNOWNS / UNKNOWNS
t = 4.0 s / x
qo = 300
vo = 60 m/s

x = (vo cos qo)t = (60 m/s)(cos 300)(4.0 s) = 208 = 210 m (due to significant figures).

Problem 24