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Sophie [7]
3 years ago
6

A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro

ps stone B with initial velocity=0.Both stones reach the river at the same time.
1-how many seconds does it take stone A to reach the river?
2-how high is the bridge?
3-what is the speed of stone A just before it strikes the river
Physics
1 answer:
Anastaziya [24]3 years ago
7 0

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.

Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12

Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

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2. A ball is dropped from rest. The acceleration due to gravity is 10m/s? and the time it
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3 0
3 years ago
The left end of a rod of length and rotational inertia is attached to a frictionless horizontal surface by a frictionless pivot,
natima [27]

Answer:

See explaination

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6 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

5 0
2 years ago
How long will be required for an object to go from a speed of 22m/s to a speed of 27m/s if the acceleration is 5.93m/s^2 ?
mario62 [17]

Answer:

Required time, t = 0.84 seconds

Explanation:

It is given that,

Initial speed of an object, u = 22 m/s

Final velocity of an object, v = 27 m/s

Acceleration, a = 5.93 m/s²

We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:

v=u+at

Where

t = time

t=\dfrac{v-u}{a}

t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}

t = 0.84 seconds

Hence, the time required for an object is 0.84 seconds.

4 0
3 years ago
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