Thank you for visiting A cannon launches a ball at a speed of 97 8 m s at an angle of 43 4 degrees How long does it take. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer:
[tex]$13.714$[/tex] seconds
Problem Solving Strategy:
- What we have here is a projectile motion problem. The equation for the y-component of an object's velocity in projectile motion is [tex]$v_y=v_{0y}-gt[/tex]. Here, [tex]v_y[/tex] is the y-component of the object's velocity. [tex]v_{0y}[/tex] is the y-component of the objects initial velocity, and [tex]g[/tex] is the gravitational constant [tex]$9.8$[/tex] [tex]m/s^2[/tex].
- I've additionally attached a useful figure below to help visualize the concepts presented.
Step 1: Find the y-component of the cannon ball's initial velocity, [tex]v_{0y}[/tex].
- [tex]v_{0y}=v_0\sin(\theta) = 97.8\sin(43.4)\approx67.197\:$ m/s[/tex]
Step 2: At the peak of the projectile motion, the y-component of the ball's velocity, [tex]v_y[/tex] will be equal to zero. Using this fact, we can solve for the time it takes to reach the peak of the parabola, which will be useful in the following steps.
- [tex]$v_y=v_{0y}-gt[/tex]
- [tex]$\Rightarrow 0=67.197-(9.8)t$[/tex]
- [tex]$\Rightarrow t=6.857[/tex] seconds
Step 3: Since the object has reached its maximum height in 6.857 seconds, we can conclude the total time is equal to [tex]2*6.857=13.714[/tex] seconds. We can see from the figure that the maximum height is reached at half of the final time. So doubling this will give us our final time.
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Rewritten by : Jeany
