Problem#1
A high-powered rifle fires a bullet with a muzzle speed of 1.00 km/s. The gun is pointed horizontally at a large bull’s eye target—a set of concentric rings—200 m away. (a) How far below the extended axis of the rifle barrel does a bullet hit the target? The rifle is equipped with a telescopic sight. It is “sighted in” by adjusting the axis of the telescope so that it points precisely at the location where the bullet hits the target at 200 m. (b) Find the angle between the telescope axis and the rifle barrel axis. When shooting at a target at a distance other than 200 m, the marksman uses the telescopic sight, placing its crosshairs to “aim high” or “aim low” to compensate for the different range. Should she aim high or low, and approximately how far from the bull’s eye, when the target is at a distance of (c) 50.0 m, (d) 150 m, or (e) 250 m? Note: The trajectory of the bullet is everywhere so nearly horizontal that it is a good approximation to model the bullet as fired horizontally in each case. What if the target is uphill or downhill? (f) Suppose the target is 200 m away, but the sight line to the target is above the horizontal by 30°. Should the marksman aim high, low, or right on? (g) Suppose the target is downhill by 30°. Should the marksman aim high, low, or right on? Explain your answers.Answer:
We use the approximation mentioned in the problem. The time to travel 200 m horizontally is
t = ∆x/vx = 200 m/(1000 m/s) = 0.20 s
The bullet falls by
∆y = vyit – ½ gt2
∆y = 0 – ½ (9.80 m/s2)(0.2 s)2 = – 0.196 m
(b) The telescope axis must point below the barrel axis by
θ = tan1 (0,196/200) = 0.05610
(c) t = ∆x/vx = 50 m/(1000 m/s) = 0.050 s. The bullet falls by only
∆y = vyit – ½ gt2
∆y = 0 – ½ (9.80 m/s2)(0.050 s)2 = – 0.0122 m
At range 50 m = ¼ x 200 m, the scope axis points to a location ½ (19,6 cm) = 4.9 cm above the
barrel axis, so the sharpshooter must aim low by 4.90 cm 1.22 cm = 3.68 cm.
(d) t = ∆x/vx = 150 m/(1000 m/s) = 0.150 s. The bullet falls by only
∆y = vyit – ½ gt2
∆y = 0 – ½ (9.80 m/s2)(0.150 s)2 = 0.110 m
At range 150 m = ¾ x 200 m, the scope axis points to a location ¾ (19,6 cm) = 14.7 cm above the
barrel axis, so the sharpshooter must aim low by 14.7 cm 11.0 cm = 3.70 cm.
(e) t = ∆x/vx = 250 m/(1000 m/s) = 0.250 s. The bullet falls by only
∆y = vyit – ½ gt2
∆y = 0 – ½ (9.80 m/s2)(0.250 s)2 = 0.306 m
At range 250 m = 5/4 x 200 m, the scope axis points to a location 5/4 (19,6 cm) = 24.4 cm above the
barrel axis, so the sharpshooter must aim high by 30.6 cm 24.4 cm = 6.12 cm.
(f), (g) Many marksmen have a hard time believing it, but they should aim low in both cases. As in case (a) above, the time of flight is very nearly 0.200 s and the bullet falls below the barrel axis by 19.6 cm on its way. The 0.0561° angle would cut off a 19.6-cm distance on a vertical wall at a horizontal distance of 200 m, but on a vertical wall up at 30° it cuts off distance h as shown, where cos 30°= 19.6 cm h , h = 22.6 cm. The marksman must aim low by 22.6 cm− 19.6 cm= 3.03 cm. The answer can be obtained by considering limiting cases. Suppose the target is nearly straight above or below you. Then gravity will not cause deviation of the path of the bullet, and one must aim low as in part (c) to cancel out the sighting-in of the telescope.
(d) t = ∆x/vx = 150 m/(1000 m/s) = 0.150 s. The bullet falls by only
∆y = vyit – ½ gt2
∆y = 0 – ½ (9.80 m/s2)(0.150 s)2 = 0.110 m
At range 150 m = ¾ x 200 m, the scope axis points to a location ¾ (19,6 cm) = 14.7 cm above the
barrel axis, so the sharpshooter must aim low by 14.7 cm 11.0 cm = 3.70 cm.
(e) t = ∆x/vx = 250 m/(1000 m/s) = 0.250 s. The bullet falls by only
∆y = vyit – ½ gt2
∆y = 0 – ½ (9.80 m/s2)(0.250 s)2 = 0.306 m
At range 250 m = 5/4 x 200 m, the scope axis points to a location 5/4 (19,6 cm) = 24.4 cm above the
barrel axis, so the sharpshooter must aim high by 30.6 cm 24.4 cm = 6.12 cm.
(f), (g) Many marksmen have a hard time believing it, but they should aim low in both cases. As in case (a) above, the time of flight is very nearly 0.200 s and the bullet falls below the barrel axis by 19.6 cm on its way. The 0.0561° angle would cut off a 19.6-cm distance on a vertical wall at a horizontal distance of 200 m, but on a vertical wall up at 30° it cuts off distance h as shown, where cos 30°= 19.6 cm h , h = 22.6 cm. The marksman must aim low by 22.6 cm− 19.6 cm= 3.03 cm. The answer can be obtained by considering limiting cases. Suppose the target is nearly straight above or below you. Then gravity will not cause deviation of the path of the bullet, and one must aim low as in part (c) to cancel out the sighting-in of the telescope.
A hawk is flying horizontally at 10.0 m/s in a straight line, 200 m above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for 2.00 seconds before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3.00 m above the ground. (a) Assuming no air resistance, find the diving speed of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse “enjoy” free fall?
Answer:
(c) We use
yf = yi + vyit – ½ gt2,
3.00 m = 200 m + 0 – ½ (9.80 m/s2)t2
–197 m = 0 – ½ (9.80 m/s2)t2
t = 6.34 s
(a) From Part (c), the raptor dives for 6.34 s − 2.00 s = 4.34 s undergoing displacement 197 m downward and (10.0 m/s)(4.34 s) = 43.4 m forward. Then the diving speed of the hawk is
v = ∆r/∆t = √[(197 m)2 + (43.4 m)2]/4.34 s
v = 46.5 m/s
θ = tan1(–197 m/43.4 m) = –77.60
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