# 3.3 Projectile motion  (Page 8/16)

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An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

(a) $\text{18}\text{.}\text{4º}$

(b) The arrow will go over the branch.

A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

Verify the ranges for the projectiles in [link] (a) for $\theta =\text{45º}$ and the given initial velocities.

$\begin{array}{}R=\frac{{{v}_{0}}^{}}{\text{sin}{2\theta }_{0}g}\\ \text{For}\phantom{\rule{0.25em}{0ex}}\theta =\text{45º},R=\frac{{{v}_{0}}^{}}{g}\end{array}$

$R=91.8\phantom{\rule{0.25em}{0ex}}\text{m}$ for ${v}_{0}=30\phantom{\rule{0.25em}{0ex}}\text{m/s}$ ; $R=163\phantom{\rule{0.25em}{0ex}}\text{m}$ for ${v}_{0}=40\phantom{\rule{0.25em}{0ex}}\text{m/s}$ ; $R=255\phantom{\rule{0.25em}{0ex}}\text{m}$ for ${v}_{0}=50\phantom{\rule{0.25em}{0ex}}\text{m/s}$ .

Verify the ranges shown for the projectiles in [link] (b) for an initial velocity of 50 m/s at the given initial angles.

The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is $6\text{.}\text{37}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{km}$ . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

(a) 560 m/s

(b) $8\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m}$

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

An arrow is shot from a height of 1.5 m toward a cliff of height $H$ . It is shot with a velocity of 30 m/s at an angle of $\text{60º}$ above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, $g$ . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

1.50 m, assuming launch angle of $45º$

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

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