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2 years ago

A ball is launched at 40m/s. find its speed at the half way up point.

1 answer:

6 0

Ek - kinetic energy
v2 - unknown speed
v1 - 40 m/s (initial speed)
Ek=1/2 mv^2
Ek in half way up is 1/2 Ek (with another V)
So, Ek in the beginning is Ek1= 1/2 mv1^2
and in half way Ek2=1/2 mv2^2
Ek1=2*Ek2
1/2 mv1^2 = 2* 1/2 mv2^2
1/2 v1^2 = v2^2
1/2 40^2 = v2^2
800 = v2^2
v2 = sqrt (800) = 28,3 m/s

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

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3 years ago

The half-life of carbon-14 is 5730 years. how long will it take for 7/8 of a sample of carbon-14 to decay

wlad13 [49]

The half life of Carbon-14 is 5730 years, how many years would it take for 7/8 of the original amount to decay?

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2 years ago

Read 2 more answers

As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a

Annette [7]

Im pretty sure it’s a because it makes more sense you know?.

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2 years ago

HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON

Schach [20]

i do not have an answer because it depends on the size and the distance lol

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2 years ago

The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The

Vika [28.1K]

Answer:

No, there won't be a collision.

Explanation:

We will use the constant acceleration formulas to calculate,

v = u + a*t

0 = 25 + (-0.1)*t

t = 250 seconds (the time taken for the passenger train to stop)

v^2 = u^2 + 2*a*s

0 = (25)^2 + 2*(-0.1)*s

s = 3125 m (distance traveled by passenger train to stop)

If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur

Speed*time = distance

Distance = (15)*(250)

Distance = 3750 m

As the distance is way more, there won’t be a collision

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2 years ago