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Angelina_Jolie [31]
3 years ago
15

A projectile is shot straight up from the earth's surface at a speed of 10,000 km/hr. how high does it go?

Physics
1 answer:
mezya [45]3 years ago
7 0
Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius. 
KE/m = (9000000/3600)^2/2 = 3858025 J/kg 
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r 
KE/m = ΔPE/m 
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m 
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)
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Answer:

The initial velocity was U=22.14m/s

Explanation:

Step one :

Applying the third equation of motion

v² = u²+ 2as

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Step two :

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Step three :

Substituting into the equation we have

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0=U²+490.5

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U=22.14m/s

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3 years ago
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is
lbvjy [14]

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

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b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

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Answer:

Momentum is conserved.

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Answer:

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