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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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3 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
4 years ago
A 817 kg car has four 8.91 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rot
KATRIN_1 [288]

Answer:

0.0107

Explanation:

We know that

The rotational kinetic energy due to four wheel is

1/2ဃ²I x 4

So

1/4mR²(v/R)² = mv²

But kinetic energy along straight path of the car is 1/2mv²

=> 1/2( 817)v ²

Kc= 408.5v²

So The fraction of total kinetic energy that is due to rotation of the wheel about their axis

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8 0
3 years ago
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Answer:

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Explanation:

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v^2 - v_0^2 = 2a\Delta s

where v0 = 0 m/s is the initial velocity of the airplane when it start accelerating, v = 161 km/h = 161*1000*(1/60)(1/60) = 44.72 m/s is the airborn speed, a = 2.6 m/s2 is the acceleration, and \Delta s is the distance of the runway, which we care looking for

44.72^2 - 0 = 2*2.6\Delta s

\Delta s = 44.72^2/(2*2.6) = 384.6 m

5 0
4 years ago
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