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UNO [17]
3 years ago
15

A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum

displacement?
Physics
1 answer:
V125BC [204]3 years ago
7 0
Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)
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A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distan
Triss [41]

Answer:

<em>The distance covered by comet is </em>200,000 km

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= \bold{\frac{distance}{time}}

Thus distance= speed*time

In this problem it is given that speed of comet= \frac{50,000km}{hr}

time travelled by the comet= 4 hours

Thus distance= speed*time

                            = 500000*4

                            = \bold{200,000km}

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Explain when a falling object is in free fall.
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4 years ago
Read 2 more answers
Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1
harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum p_{i} (before the elastic collision) must be equal to the final momentum p_{f} (after the elastic collision):

p_{i}=p_{f} (1)

Being:

p_{i}=m_{1}V_{i} + m_{2}U_{i}

p_{f}=m_{1}V_{f} + m_{2}U_{f}

Where:

m_{1}=200 kg +100 kg=300 kg is the combined mass of Tubby and Libby with the car

V_{i}=10 m/s is the velocity of Tubby and Libby with the car before the collision

m_{2}=25 kg + 100 kg=125 kg is the combined mass of Flubby with its car

U_{i}=0 m/s is the velocity of Flubby with the car before the collision

V_{f}=4.12 m/s is the velocity of Tubby and Libby with the car after the collision

U_{f} is the velocity of Flubby with the car after the collision

So, we have the following:

m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f} (2)

Finding U_{f}:

U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}} (3)

U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg} (4)

Finally:

U_{f}=14.1 m/s

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