Problem#30
The determined coyote is out once more in pursuit of the elusive roadrunner. The coyote wears a pair of Acme jetpowered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2 (Fig. 1. The coyote starts at rest 70.0 m from the brink of a cliff at the instant the roadrunner zips past him in the direction of the cliff. (a) If the roadrunner moves with constant speed, determine the minimum speed he must have in order to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. His skates remain horizontal and continue to operate while he is in flight, so that the coyote’s acceleration while in the air is (15.0ˆi – 9.80ˆj ) m/s2. (b) If the cliff is 100 m above the flat floor of a canyon, determine where the coyote lands in the canyon. (c) Determine the components of the coyote’s impact velocity. |
| Fig.1 |
Answer:
(a) For Coyote:
∆x = ½ at2
70.0 m = ½ (15.0 m/s2)t2
t = 3.06 s
Roadrunner:
∆x = vit
70.0 m = vi(3.06 s)
vi = (70.0 m)/(3.06 s) = 22.9 m/s
(b) At the edge of the cliff,
vxi = at = (15.0 m/s)(3.06 s) = 45.8 m/s
Substituting into
∆y = ½ ayt2
We find,
–100 m = ½ (9.80 m/s2)t2
t = 4.52 s
then,
∆x = vxit + ½ axt2
∆x = (45.8 m/s)(4.52 s) + ½ (15.0 m/s)(4.52 s)2 = 360.0 m
(c) For the Coyote’s motion through the air
vxf = vxi + axt = 45.8 m/s + (15 m/s2)(4.52 s) = 114 m/s
vyf = vyi + ayt = 0 – (9.8 m/s2)(4.52 s) = –44.3 m/s
Problem#2
A skier leaves the ramp of a ski jump with a velocity of 10.0 m/s, 15.0° above the horizontal, as in Figure 2. The slope is inclined at 50.0°, and air resistance is negligible. Find (a) the distance from the ramp to where the jumper lands and (b) the velocity components just before the landing. (How do you think the results might be affected if air resistance were included? Note that jumpers lean forward in the shape of an airfoil, with their hands at their sides, to increase their distance. Why does this work?)
 |
| Fig.2 |
Answer:
Given by:
vxi = vi cos 15.00 = (10.0 m/s) cos 15.00 = 9.66 m/s
vyi = vi sin 15.00 = (10.0 m/s) sin 15.00 = 2.59 m/s
Given by:
vxi = vi cos 15.00 = (10.0 m/s) cos 15.00 = 9.66 m/s
vyi = vi sin 15.00 = (10.0 m/s) sin 15.00 = 2.59 m/s
 |
| Fig.3 |
Equations of motion
x-axis:
∆x = x = vxit
D cos 50.00 = (9.66 m/s)t (*)
and
y-axis:
∆y = vyit – ½ gt2
–D sin 50.00 = (2.59 m/s)t – ½ (9.80 m/s2)t2
from (*) and (**), we get
–tan 50.00 (9.66 m/s)t = (2.59 m/s)t – ½ (9.80 m/s2)t2
4.9t2 – 14.10t = 0
t = 0 or t = 2.88 s
then from (*), we get
D cos 50.00 = (9.66 m/s)(2.88 s)
D = 43.25 m
(b) since ax = 0,
vxf = vxi = 9.66 m/s
vyf = vyi – gt = 2.59 m/s – (9.80 m/s2)(2.88 s) = 25.63 m/s
Air resistance would decrease the values of the range and maximum height. As an airfoil, he
can get some lift and increase his distance.
Problem#3
An enemy ship is on the east side of a mountain island, as shown in Figure 4. The enemy ship has maneuvered to within 2500 m of the 1 800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
∆x = x = vxit
D cos 50.00 = (9.66 m/s)t (*)
and
y-axis:
∆y = vyit – ½ gt2
–D sin 50.00 = (2.59 m/s)t – ½ (9.80 m/s2)t2
from (*) and (**), we get
–tan 50.00 (9.66 m/s)t = (2.59 m/s)t – ½ (9.80 m/s2)t2
4.9t2 – 14.10t = 0
t = 0 or t = 2.88 s
then from (*), we get
D cos 50.00 = (9.66 m/s)(2.88 s)
D = 43.25 m
(b) since ax = 0,
vxf = vxi = 9.66 m/s
vyf = vyi – gt = 2.59 m/s – (9.80 m/s2)(2.88 s) = 25.63 m/s
Air resistance would decrease the values of the range and maximum height. As an airfoil, he
can get some lift and increase his distance.
Problem#3
An enemy ship is on the east side of a mountain island, as shown in Figure 4. The enemy ship has maneuvered to within 2500 m of the 1 800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
 |
| Fig.4 |
Answer:
Find the highest firing angle θH for which the projectile will clear the mountain peak; this will yield the range of the closest point of bombardment. Next find the lowest firing angle; this will yield
the maximum range under these conditions if both θ H and θL are > 450 ; x = 2500 m, y = 1800 m,
vi = 250 m/s.
We have:
x-axis: xf = xi + vxit or t = xf/(vi cos θ)
y-axis:
yf = yi + vyit – ½ gt2
yf = 0 + (vi sin θ)(xf/(vi cos θ)) – ½ g[xf/(vi cos θ)]2
yf = xf tan θ – gxf2/(2vi2 cos2 θ)
but 1/cos2 θ = 1 + tan2 θ, so
yf = xf tan θ – (gxf2/2vi2)(1 + tan2 θ)
with
gxf2/2vi2 = (9.80 m/s2)(2500 m)2/[2(250 m/s)2] = 490
1800 = 2500 tanθ – 490(1 + tan2 θ)
49 tan2 θ – 250 tan θ + 229 = 0
tan θ12 = {250 ± √[(-250)2 – 4 x 49 x 229]}/(2 x 49)
tan θ = 3.905 or 1.197
so we get
θL = 75.60 and θH = 50.10
we have Range, R = vi2 sin 2θ/g
Range (at θH) = (250 m/s)2 sin (2 x 50.10)/(9.80 m/s2) = 3070 m from enemy ship and
3070 m – 2500 m – 300 m = 270 m from shore.
Range (at θL) = (250 m/s)2 sin (2 x 75.60)/(9.80 m/s2) = 6280 m from enemy ship and
6280 m – 2500 m – 300 m = 3480 m from shore.
Therefore, safe distance is < 270 m or > 3480 m from the shore.
Find the highest firing angle θH for which the projectile will clear the mountain peak; this will yield the range of the closest point of bombardment. Next find the lowest firing angle; this will yield
the maximum range under these conditions if both θ H and θL are > 450 ; x = 2500 m, y = 1800 m,
vi = 250 m/s.
We have:
x-axis: xf = xi + vxit or t = xf/(vi cos θ)
y-axis:
yf = yi + vyit – ½ gt2
yf = 0 + (vi sin θ)(xf/(vi cos θ)) – ½ g[xf/(vi cos θ)]2
yf = xf tan θ – gxf2/(2vi2 cos2 θ)
but 1/cos2 θ = 1 + tan2 θ, so
yf = xf tan θ – (gxf2/2vi2)(1 + tan2 θ)
with
gxf2/2vi2 = (9.80 m/s2)(2500 m)2/[2(250 m/s)2] = 490
1800 = 2500 tanθ – 490(1 + tan2 θ)
49 tan2 θ – 250 tan θ + 229 = 0
tan θ12 = {250 ± √[(-250)2 – 4 x 49 x 229]}/(2 x 49)
tan θ = 3.905 or 1.197
so we get
θL = 75.60 and θH = 50.10
we have Range, R = vi2 sin 2θ/g
Range (at θH) = (250 m/s)2 sin (2 x 50.10)/(9.80 m/s2) = 3070 m from enemy ship and
3070 m – 2500 m – 300 m = 270 m from shore.
Range (at θL) = (250 m/s)2 sin (2 x 75.60)/(9.80 m/s2) = 6280 m from enemy ship and
6280 m – 2500 m – 300 m = 3480 m from shore.
Therefore, safe distance is < 270 m or > 3480 m from the shore.
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