Your Frisbee is stuck in a tree at a height of 9 m above you. You throw a baseba

Question

Your Frisbee is stuck in a tree at a height of 9 m above you. You throw a baseball up toward it to try to knock it loose.

(a) First you try to stand directly below the Frisbee and throw the baseball straight up. With what minimum speed must you throw the ball to just barely reach the Frisbee?

(b) How long does it take the baseball to reach the Frisbee when you throw it at this speed?

(c) You find that it is difficult for you to throw a baseball straight up at this speed, so you back up a horizontal distance of 3 m. You throw the baseball with a velocity whose y component equals the answer from part (a), so that it barely reaches the correct height. What is the x component of the velocity needed to hit the Frisbee?

(d) With what speed and with what angle above the horizontal must you throw the ball to obtain the velocity you calculated in part (c)?

Explanation / Answer

a)

We use the equation of motion,

v^2 = u^2 + 2as

In vertical motion,

v = final velocity = 0

u = initial velocity

a = -9.8 m/s2

s = 9

So, 0^2 = u^2 - 2*9.8*9

So, u = 13.3 m/s <--------answer

b)

Now, using the equation of motiom,

v = u +at

So, 0 = 13.3 -9.8*t

So, t = 1.36 s <--------answer

c)

y- component of u', u'y = 13.3 m/s

So, distance traveled in x-component , d = u'x*t = 3m

where u'x = x-component of velocity

So, u'x = 3/1.36 = 2.21 m/s <----------answer

d)

So, the speed = sqrt(13.3^2 + 2.21^2) = 13.5 m/s

angle = atan(13.3/2.21) = 80.6 deg <------answer

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