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Katen [24]
3 years ago
8

A robot on a basketball court needs to throw the ball in the hoop. The ball have a velocity of 10 m/s and an an angle of 60 degr

ees. The high of hoop is 0.75 meters and the horizontal distance between the robot and hoop is 5 meters. A) draw a digram.
B)What is the high that the robot needs to throw the ball from.

Physics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:

Initial velocity = 10 m/s

θ = 60°

This is the case of projectile motion

So the horizontal component of velocity 10 m/s  =  10 cosθ

u = 10 cosθ

u = 10 cos 60°

u=5 m/s

x= 5 m

So in the horizontal direction

x = u .t

5 = 5 .t

t = 1 sec  The vertical component of velocity 10 m/s = 10 sinθ

Vo= 10 sinθ

Vo= 10 sin 60°

Vo = 8.66 m/s

V^2=V_o^2+2gh

0=V_o^2-2gh

0=8.66^2-2\times 10\times h

h=3.75 m

So height of robot = 3.75 - 0.75 m

   height of robot =3 m

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Answer:

To achieve the velocity of 40 m/sec height will become 4 times  

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is \frac{1}{2}mv^2

So total energy at the top of the hill =mgh+0=mgh

At the bottom of the hill kinetic energy is equal to \frac{1}{2}mv^2 and potential energy will be 0

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Form energy conservation mgh=\frac{1}{2}mv^2

v=\sqrt{2gh}, for v = 20 m/sec

20=\sqrt{2\times 9.8\times h}

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19.6h=400

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Now if velocity is 0 m/sec

40=\sqrt{2gh}

19.6h=1600

h = 81.63 m

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q = (120 g) (0.40 J/g/K) (40°C − 0°C)

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Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

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Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

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