Question

In Problems 78, 79, and 80 you are given the equations that are used to solve a problem. For each of these, you are to write a realistic problem for which these are the correct equations. Be sure that the answer your problem requests is consistent with the equations given.﻿100 m=0 m+(50cos⁡θ m/s)t10 m=0 m+(50sin⁡θ m/s)t1−12(9.80 m/s2)t12\begin{aligned}
100 \, 	ext{m} &= 0 \, 	ext{m} + (50 \cos 	heta \, 	ext{m/s}) $$t_1$$ \
0 \, 	ext{m} &= 0 \, 	ext{m} + (50 \sin 	heta \, 	ext{m/s}) $$t_1$$ - \frac{1}{2} (9.80 \, 	ext{m/s}^2) $$t_1^2$$
\end{aligned}100m0m​=0m+(50cosθm/s)t1​=0m+(50sinθm/s)t1​−21​(9.80m/s2)t12​​

Accepted Answer

Step 1: Analyze the given equations. The first equation, 100 m = 0 m + (50cos⁡θ m/s)t₁, represents the horizontal motion of a projectile, where the horizontal displacement is 100 m, the initial horizontal velocity is (50cosθ) m/s, and t₁ is the time of flight. The second equation, 0 m = 0 m + (50sin⁡θ m/s)t₁ - (1/2)(9.80 m/s²)t₁², represents the vertical motion of the projectile, where the vertical displacement is 0 m (it returns to the same vertical level), the initial vertical velocity is (50sinθ) m/s, and the acceleration due to gravity is -9.80 m/s². Step 2: Recognize the type of problem. These equations describe the motion of a projectile launched at an angle θ with an initial velocity of 50 m/s. The horizontal displacement is 100 m, and the projectile lands back at the same vertical level from which it was launched. The problem likely involves finding the angle θ or the time of flight t₁. Step 3: Set up a realistic problem. A possible problem could be: 'A ball is launched from the ground with an initial velocity of 50 m/s at an angle θ above the horizontal. The ball travels a horizontal distance of 100 m before landing back on the ground. Determine the angle θ at which the ball was launched and the time of flight t₁.' Step 4: Solve for time of flight t₁ using the horizontal motion equation. Rearrange the first equation to isolate t₁: t₁ = 100 m / (50cosθ m/s). This equation relates the time of flight to the horizontal velocity component and the horizontal displacement. Step 5: Substitute t₁ into the vertical motion equation to solve for θ. Replace t₁ in the second equation with the expression derived in Step 4. This will result in an equation involving θ that can be solved using trigonometric identities or numerical methods. The solution will yield the angle θ that satisfies both the horizontal and vertical motion equations.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Projectile Motion

Trigonometric Functions

Kinematic Equations