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Chapter 3: Problem 47

A clump of soft clay is thrown horizontally from \(8.50 \mathrm{m}\) above the ground with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) Where is the clay after 1.50 s? Assume it sticks in place when it hits the ground.

Short Answer

Expert verified
Answer: After 1.50 seconds, the clay projectile is located at a horizontal distance of 30.0 meters and a vertical height of 6.21 meters above the ground.

Step by step solution

01

Determine the horizontal distance

To find the horizontal distance covered by the clay during the given time, we use the following equation for horizontal motion: \(x = x_0 + v_x t\) Where: \(x\) - horizontal distance after 1.50 s \(x_0\) - initial horizontal distance (0, as the projectile is thrown from the starting point) \(v_x\) - horizontal velocity (20.0 m/s) \(t\) - time (1.50 s) Plugging in the values, we get: \(x = 0 + 20.0 \times 1.50\) \(x = 30.0\) meters So, the clay moved 30.0 meters horizontally.
02

Determine the vertical distance

To find the vertical distance covered by the clay during the given time, we use the following equation for vertical motion: \(y = y_0 + v_y t - \frac{1}{2}gt^2\) Where: \(y\) - vertical distance after 1.50 s \(y_0\) - initial vertical distance (8.50 m) \(v_y\) - initial vertical velocity (0, as the projectile is thrown horizontally) \(g\) - acceleration due to gravity (9.81 m/s²) \(t\) - time (1.50 s) Plugging in the values, we get: \(y = 8.50 - \frac{1}{2} \times 9.81 \times (1.50)^2\) \(y \approx 2.294\, 25\) meters So, the clay has moved 2.29 meters vertically (downward).
03

Check if the clay has hit the ground

Since the projectile was thrown from a height of 8.50 meters, we can check whether it has hit the ground after 1.50 seconds of flight: Initial height - vertical distance after 1.50 seconds = remaining height \(8.50 - 2.294\, 25 = 6.205\, 75\) meters Since there is still remaining height between the clay and the ground, it hasn't hit the ground yet.
04

State the final position of the clay

Since the clay has not hit the ground yet, we use the horizontal and vertical distances to determine its position. The clay has moved 30.0 meters in the horizontal direction and 2.29 meters in the vertical direction downward: Horizontal position: 30.0 meters Vertical position: 6.21 meters above the ground (rounded to two decimal places) So, after 1.50 seconds, the clay is located at a horizontal distance of 30.0 meters and a vertical height of 6.21 meters above the ground.

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Most popular questions from this chapter

You are serving as a consultant for the newest James Bond film. In one scene, Bond must fire a projectile from a cannon and hit the enemy headquarters located on the top of a cliff \(75.0 \mathrm{m}\) above and \(350 \mathrm{m}\) from the cannon. The cannon will shoot the projectile at an angle of \(40.0^{\circ}\) above the horizontal. The director wants to know what the speed of the projectile must be when it is fired from the cannon so that it will hit the enemy headquarters. What do you tell him? [Hint: Don't assume the projectile will hit the headquarters at the highest point of its flight. \(]\)
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