Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 2, Problem 62P

(a)

To determine

Maximum height that can be reached by stone.

(a)

Expert Solution
Check Mark

Answer to Problem 62P

Maximum height is 21.1m.

Explanation of Solution

Initial speed is 19.6m/s and the initial height of stone from the ground is 1.50m.

Write the Newton’s equation for an object moving vertically upwards.

vf2vi2=2g(yfyi)

Here, the velocity at the highest point of motion is vf, initial velocity of motion is vi, gravitational acceleration is g, final position from the ground is yf, and the initial position from the ground is yi.

Conclusion:

Substitute 0m/s for vf, 19.6m/s for vi, 9.8m/s2 for g, and 1.50m for yi in the above equation to find yf.

(0m/s)2(19.6m/s)2=2(9.8m/s2)(yf1.50m)384m2/s2=19.6m/s2(yf1.50m)

Rewrite the above relation in terms of yf.

yf=384m2/s219.6m/s2+1.50m=21.1m

(b)

To determine

Total time taken by the stone for its whole journey before reaching the ground.

(b)

Expert Solution
Check Mark

Answer to Problem 62P

Time taken is 4.08s.

Explanation of Solution

Initial speed is 19.6m/s and the initial height of stone from the ground is 1.50m.

Write the equation for time taken by the stone to reach the maximum height.

tup=2(yfyi)g (I)

Here, the time taken by stone for its upward motion is tup.

Write the equation for time taken by the stone for its downward motion.

tdown=2yfg (II)

Here, the time taken by stone for its downward motion is tdown.

Write the equation for total time taken by stone for its whole travelling before reaching he ground.

t=tup+tdown (III)

Here, the total time taken is t.

Conclusion:

Substitute 9.8m/s2 for g, 21.1m for yf and 1.50m for yi in equation (I) to find tup.

tup=2(21.1m1.50m)(9.8m/s2)=2.0s

Substitute 9.8m/s2 for g and 21.1m for yf a in equation (II) to find tdown.

tdown=2(21.1m)9.8m/s2=2.08s

Substitute 2.0s for tup and 2.08s for tdown in equation (III) to find t.

t=2.0s+2.08s=4.08s

Therefore, the time taken is 4.08s.

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