A circus performer stands on a platform and throws an apple from a height of 49 m above the ground with an initial velocity
v0 as shown in the figure below. A second, blindfolded performer must catch the apple. If
v0 = 30 m/s,
how far from the end of the platform should the second performer stand? (Assume θ = 30°.) the answer key is 51.5m. If you need to see the figure , give me your email and I will send it to you as fast as I could
3 answers
no thanks
Okay Anonymous, I don't know who you are
but you don't need to see the figure then show me how to solve the problem. Your resonse doesn't help in anyway on solving this question.
but you don't need to see the figure then show me how to solve the problem. Your resonse doesn't help in anyway on solving this question.
1st:-1/2*9.81t^2+(30*sin(-30)t)+49-->-4.9t^2-15t+49
2nd: use the quadratic formula to find time in second-->(-(-15)+or -square root (-15-4(-4.9*49))divide the whole thing by -9.81=-5.04s or 1.98s
3rd: take 1.98s multiply for the answer--> 30cos(-30)*1.98=51.5m
2nd: use the quadratic formula to find time in second-->(-(-15)+or -square root (-15-4(-4.9*49))divide the whole thing by -9.81=-5.04s or 1.98s
3rd: take 1.98s multiply for the answer--> 30cos(-30)*1.98=51.5m
6 answers