An Object is projected from the ground with an initial velocity of 45 km/h at 30 degrees above the horizontal. Find the maximum height attained by the object.
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A: Given,initial velocity u=40.003 ft/s=12.193 m/s
Q: How to solve for the magnitude of average velocity.
A: Velocity = dsdt Average velocity = total displcement covertotal time
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Q: x = 3t3 + 4t + 5.
A: Refer attachment
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Q: whats the definition of average distance?
A: Distance is the length of space between the starting and the ending point.
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A: Given: v1=20 m/st1=2 min = 120 sv2=30 m/st2=3 min = 180sv3=15 m/st3=1 min=60s
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