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Chapter 3: Problem 50

A projectile is launched at a \(60^{\circ}\) angle above the horizontal on level ground. The change in its velocity between launch and just before landing is found to be \(\Delta \vec{v}=\vec{v}_{\text {landing }}-\vec{v}_{\text {launch }}=-20 \hat{y} \mathrm{~m} / \mathrm{s}\). What is the initial velocity of the projectile? What is its final velocity just before landing?

Short Answer

Expert verified
Answer: The final velocity of the projectile just before landing is given by \(\vec{v}_{landing} = \left( v_{0}\cos(60^{\circ})\right)\hat{x} + \left(v_{0}\sin(60^{\circ}) -20 \,\text{m/s}\right)\hat{y}\), where \(v_{0}\) is the initial velocity of the projectile.

Step by step solution

01

Determine the vertical and horizontal components of the change in velocity

The change in velocity is given as \(\Delta \vec{v} = -20\hat{y}\,\text{m/s}\). Since the projectile is launched on level ground and lands on level ground, there is no horizontal change in velocity, so \(\Delta v_x = 0\). The vertical change in velocity is given as \(\Delta v_y = -20\,\text{m/s}\).
02

Write the projectile motion equations for vertical and horizontal components

We can write the equations for the vertical and horizontal motion components. Horizontal motion: \(v_{x} = v_{0x}\) Vertical motion: \(v_{y} = v_{0y} - gt\) where \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(9.8\,\text{m/s}^2\)) and \(t\) is the time of flight.
03

Find the initial vertical velocity component from the change in vertical velocity

Using the equation for vertical motion, we can relate the change in vertical velocity to the initial velocity, \(v_y - v_{0y} = -gt\) Since we know the change in vertical velocity, we can write: \(v_{0y} - (-20\,\text{m/s}) = -gt\) \(v_{0y} = -20\,\text{m/s} + gt\)
04

Use the launch angle to find the initial horizontal velocity component

We know the projectile is launched at a \(60^{\circ}\) angle above the horizontal, so we can find the initial horizontal velocity component as: \(v_{0x} = v_{0} \cos(60^{\circ})\) Where \(v_{0}\) is the initial velocity of the projectile.
05

Use the launch angle to find the initial vertical velocity component

Similarly, we can find the initial vertical velocity component using the same angle: \(v_{0y} = v_{0} \sin(60^{\circ})\)
06

Equating initial vertical velocity components to find the time of flight

From step 3 and step 5, we have, \(v_{0} \sin(60^{\circ}) = -20\,\text{m/s} + gt\) Solve for \(t\): \(t = \frac{v_{0} \sin(60^{\circ}) +20\,\text{m/s}}{g}\)
07

Find the initial velocity of the projectile

Now, we can square both horizontal and vertical components of initial velocity and sum them to find the initial velocity: \(v_{0}^2 = v_{0x}^2 + v_{0y}^2 \) \(v_{0}^2 = \left(v_{0}\cos(60^{\circ})\right)^2 + \left(v_{0}\sin(60^{\circ})\right)^2\) Solve for \(v_{0}\): \(v_{0} = \sqrt{\left(v_{0}\cos(60^{\circ})\right)^2 + \left(v_{0}\sin(60^{\circ})\right)^2}\)
08

Calculate the final velocity just before landing

To calculate the final velocity just before landing, we account for the fact that the horizontal component remains unchanged throughout the flight and the vertical component has changed as given: \(\vec{v}_{landing} = \left( v_{0}\cos(60^{\circ})\right)\hat{x} + \left(v_{0}\sin(60^{\circ}) -20 \,\text{m/s}\right)\hat{y}\)

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