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Elenna [48]
3 years ago
5

A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high d

oes it go
Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

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4 0
4 years ago
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie
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Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

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Momentum is given as p=mv

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0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

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Let distance move by the small mass be m

Work done by M mass

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Using conservation of energy

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Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

7 0
3 years ago
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