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TEA [102]
3 years ago
6

A cannonball of mass 15 kg is fired vertically upward from a cannon with an initial speed of 20 m/s. The cannonball travels upwa

rd, reaches a maximum height of 40 m, and returns to the ground. Now you decide to shoot a 30 kg cannonball vertically upward using the same cannon. If we assume that all cannonballs leave this cannon with an initial speed of 20 m/s, what maximum height does the 30 kg cannonball reach
Physics
1 answer:
NemiM [27]3 years ago
8 0

Answer:

30kg ball will have equal traveling height with the 15kg

Explanation:

When air resistance is negligible, the mass of the object does not affect the maximum height or the time spent in the air.

So if mass does not affect traveling height when sir resistance is neglected...the 30kg ball will have equal traveling height with the 15kg ball provided they have equall velocities and equal angle if projection.

In this case it has 20 m/s and they are all projected vertically.

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A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

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3 years ago
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