In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a
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Answer:
Maximum height of the ball, h(t) = 27.56 m
Explanation:
It is given that, a ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec.
The height of the ball as a function of time t is given by :
h₀ is initial height, h₀ = 0
So, .........(1)
For maximum/minimum height,
...(2)
t = 1.31 s
Differentiating equation (2) wrt t
h''(t) = -32 < 0
So, at t = 1.31 seconds we will get the maximum height.
Put the value of t in equation (1)
h(t) = 27.56 m
Hence, this is the required solution.