Question

3. In a carnival booth, you win a stuffed giraffe if you tossed a coin into a small
dish. The dish is on the shelf above the point where the quarter leaves your
hand and is a horizontal distance of 2.1 m from this point. If you toss the coin
with a velocity of 6.4m/s at an angle of 600 above the horizontal, the coin
lands in the dish. You can ignore air resistance.
64 -
(a) What is the height of the shelf above the point where the coin leaves your
hand?
(b) What is the vertical component of the velocity of the coin just before it lands
in the dish?
4. A test rocket is launched by accelerating it along a 200.0 m incline at 1.25
m/s starting from rest at point A (Fig P3.47). The incline rises at 35.00 above​

Answer 1

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Answer 2

The answers are 1.49 m and 3.34 m/s.

Given:

The horizontal distance of the dish = 2.1 m

The velocity of throw = 6.4 m/s

The angle of throw = 60°

To Find:
Height of the shelf

The vertical component of the velocity of the coin just before it lands in the dish

Solution:
We know that the equation of motion for a projectile motion is given by

$y=x\tan\theta(1-\frac{x}{R} )$

The formula for the range of the projectile R is

$R = \frac{u^2*\sin2\theta}{g}$

We know the given quantities

$x=2.1 m\\u=6.4\\\theta=60$

Hence using these values we get

$R = 3.55\\\\y=\sqrt{3}*2.1(1-\frac{2.1}{3.55} )\\\\y=1.49 m$

Using the velocity relation in the vertical direction,

$v^2=u^2-2gy\\\\v=\sqrt{6.4^2-2*10*1.49} \\\\v=3.34 m/s$

Hence the height of the shelf is 1.49 m and the vertical component of the velocity of the coin just before it lands in the dish is 3.34 m/s.

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