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Chapter 3: Problem 53

A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is \(40 \mathrm{m} / \mathrm{s}\) at an angle of \(37^{\circ}\) with respect to the horizontal and the cannonball is $7.0 \mathrm{m}$ above the ground at this time. (a) What is the maximum height above the ground reached by the cannonball? (b) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land? (c) What are the \(x\) - and y-components of the cannonball's velocity just before it lands? The \(y\) -axis points up.

Short Answer

Expert verified
Question: Calculate the maximum height above the ground, the horizontal distance from the release point when the cannonball lands, and the x- and y-components of the cannonball's velocity just before it lands, given that its initial velocity is 40 m/s at an angle of 37° with an initial height of 7 m above the ground.

Step by step solution

01

Determine the components of initial velocity.

Given that the initial velocity (\(v_0\)) of the cannonball is \(40 \mathrm{m} / \mathrm{s}\) at an angle of \(37^{\circ}\), we need to calculate the horizontal (\(v_{0x}\)) and vertical components (\(v_{0y}\)) of that velocity. Using the equations: \(v_{0x} = v_0 \cos (\theta)\) \(v_{0y} = v_0 \sin (\theta)\) Where \(\theta = 37^{\circ}\).
02

Step 2:Compute the time to reach the maximum height (\(t_{max\_height}\)).

At the maximum height, the vertical velocity (\(v_y\)) will be zero, so we can use the following equation to find the time: \(v_y = v_{0y} - gt\) Where \(g = 9.81 \mathrm{m}/\mathrm{s}^{2}\) is the acceleration due to gravity. Setting \(v_y = 0\) and solving for \(t\) we get: \(t_{max\_height} = \frac{v_{0y}}{g}\)
03

Calculate the maximum height above the ground (\(h_{max}\)).

To find the maximum height above the ground, we will use the equation: \(h_{max} = y_0 + v_{0y}t_{max\_height} - \frac{1}{2}gt_{max\_height}^2\) where \(y_0 = 7.0 \mathrm{m}\) is the initial height above the ground.
04

Compute the time of flight (\(t_{total}\)).

To find the time it takes for the cannonball to hit the ground, we can use the equation: \(y = y_0 + v_{0y}t - \frac{1}{2}gt^2\) Setting \(y = 0\), we can solve for time (\(t_{total}\)), which is the time of flight.
05

Calculate the horizontal distance from the release point when it lands.

To find the horizontal distance from the release point, we will use the equation: \(x = x_0 + v_{0x}t\) Where \(x_0 = 0\) is the initial horizontal position and \(t = t_{total}\), the time of flight.
06

Compute the \(x\)- and \(y\)-components of the cannonball's velocity just before it lands.

To find the velocities just before landing, we can use the equations: \(v_x = v_{0x}\) \(v_y = v_{0y} - gt_{total}\) With all these equations and steps, we can find the maximum height above the ground, the horizontal distance from its release point when it lands, and the \(x\)- and \(y\)- components of the cannonball's velocity just before it lands.

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