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egoroff_w [7]
3 years ago
11

A tennis ball is thrown vertically upward with an initial velocity of +6.2 m/s. What will the ball’s velocity be when it returns

to its starting point? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s.
Physics
1 answer:
anastassius [24]3 years ago
4 0

Answer:

v_{f}=-6.2 m/s

Explanation:

The ball will rise decreasing its speed until it reaches the highest point where its speed will be zero. From this point the tennis ball will begin to fall again, in the free fall the tennis ball will gain speed but now in the opposite direction. When it returns to the same point where it was launched, its speed will be the same as the one that was launched but with the opposite sign.

v_{f}=-6.2 m/s

We can check this using the equation:

v_{f}^2=v_{i}^2+2gh

where v_{i}=+6.2 m/s

ang h is the height, but because the ball returns to the same point where it started, h =0

then

v_{f}^2=v_{i}^2

v_{f}=v_{i}

the initial and final velocity will be the same in number, but we know that the ball is going in the opposite direction, so the final velocity must have the opposite sign from the initial velocity

so if v_{i}=+6.2 m/s,

v_{f}=-6.2 m/s

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kotegsom [21]

Answer:

h = 11.47 m

Explanation:

Initial speed pf the t-shirt gun is 15 m/s

We need to find the maximum distance covered by the t-shirt. It is based on the conservation of energy. The maximum distance covered is given by :

h=\dfrac{u^2}{2g}\\\\h=\dfrac{(15)^2}{2\times 9.8}\\\\h=11.47\ m

So, it will cover a distance of 11.47 m.

7 0
3 years ago
Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight dow
bulgar [2K]

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

F=\frac{\Delta p}{\Delta t}

where

F is the force exerted

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be rewritten as

\Delta p = m(v-u)

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

F=\frac{m(v-u)}{\Delta t}

In this problem we have:

\frac{m}{\Delta t}=0.030 kg/s is the mass rate

u=-15 m/s is the initial velocity

v=+15 m/s is the final velocity

Therefore, the force exerted by the hail on the roof is:

F=(0.030)(+15-(-15))=0.9 N

6 0
3 years ago
A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th
Zolol [24]

Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we  apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we  have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.

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