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
o = 0
vo = 18.0 m/s
y = yo + (vo sino )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
o = 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o) = 18.0 m/s
vy = (vo sino) – 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
o = 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
0
o = 53.0
vo = 20.0 m/s
yo = 0
y = yo + (vo sino )t – ½ gt2
But we do not know what t is so we need to find t first:
x = (vo coso)t
36.0 m = (20.0 m/s)(cos 53.0)t
t = 2.99 s
y = yo + (vo sino )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 o) – 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 30 0 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 ydirection is equal to ZERO.
KNOWNS
UNKNOWNS
t = 4.0 s
v
0
o = 30
vo = 60 m/s
vy = 0
vy = 0
vx = (vo cos o) = (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
0
o = 30
vo = 60 m/s
x = (vo cos o)t = (60 m/s)(cos 300)(4.0 s) = 208 = 210 m (due to significant figures).
Problem 24