A circus performer stands on a platform and throws an apple from a height of 50 m above the ground with an initial velocity

Question

v0
as shown in the figure below. A second, blindfolded performer must catch the apple. If
v0 = 21 m/s,
how far from the end of the platform should the second performer stand? (Assume θ = 30°.) The answer is 41.8. Can somebody shows me step by step to the answer

Answer 1

The answer key is 41.8 but I don't know how to solve for this question so that I can get the same answer.

Answer 2

The height of the ball is given by

h(t) = 50 + 21/2 t - 4.9t^2

Assuming the catcher is on the ground, you need

h(t) = 0, so t=4.44

So. Now you know that horizontal speed of the apple is

21 cos30° = 18.186 m/s

So, it will travel 4.44 * 18.186 = 80.75 meters.

Hmmm. Either the length of the platform is 38.95 meters long, or the catcher is not on the ground.

Have I misinterpreted the situation, having no diagram?

Answer 3

Is there anyway for I to send you the picture of the figure so that you can have a better understanding and help me with this question

Answer 4

1st:use this format equation--> -1/2*9.81t^2+(21*sin(-30)t)+50 which gives -->-4.9t^2-10.5t+50
2nd: now use the quadratic formula to find the time-->(-(-10.5)+/- the square root of(-10.5^2-(4*-4.9*50)))divide by(2*-4.9)=-4.44 and2.30
3rd: use only the positive value 2.30 and multiply with 21cos(-30)--> 21cos(-30)*(2.30)=41.8m

Answer 5

Lmaoo

Answer 6

Lma0