You shoot an arrow with a mass of 0.54 kg at 45^o above the horizontal. The bow exerts a force of 125 N for 0.65 s. With no air resistance, the maximum height the arrow reaches is A) 1.2 km. B) 5.4 m. C) 0.57 km. D) 0.29 km. E) 0.61 km.

Question

Quizplus · Accepted Answer

First, we need to determine the initial velocity of the arrow when it leaves the bow. We can use the formula:

F = ma

where F is the force exerted by the bow, m is the mass of the arrow, and a is the acceleration of the arrow. Rearranging this formula to solve for a, we get:

a = F/m

a = 125 N / 0.54 kg

a = 231.48 m/s^2

This acceleration is only applied for 0.65 s, so we can use the kinematic equation:

y = yo + vot + (1/2)at^2

where y is the final height, yo is the initial height (0), vo is the initial velocity (which we want to solve for), a is acceleration (231.48 m/s^2), and t is the time (0.65 s). We are interested in the maximum height, so we want the final velocity at this point to be 0.

0 = vo + at

vo = -at

vo = -231.48 m/s^2 * 0.65 s

vo = -150.5 m/s

Note that we obtained a negative value for vo, which indicates that the arrow is moving in the opposite direction of the bow's force. This is because we chose upward as the positive direction, but the arrow will actually travel downward before reaching its maximum height.

Now, we can find the maximum height using the formula:

y = yo + vot + (1/2)at^2

y = 0 + (-150.5 m/s) * t + (1/2) * (-9.8 m/s^2) * t^2

At the maximum height, the arrow will have zero vertical velocity, so we can set:

0 = -150.5 m/s + (-9.8 m/s^2) * t

t = 15.36 s

Now we can substitute t back into the equation to find the maximum height:

y = 0 + (-150.5 m/s) * 15.36 s + (1/2) * (-9.8 m/s^2) * (15.36 s)^2

y = 566.3 m

Therefore, the maximum height the arrow reaches is 0.57 km.

You Shoot an Arrow with a Mass of 0