The two dry sand samples are examined by changing one parameter. The independent variable, the one that is intentionally manipul
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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.
To solve this problem it is necessary to apply the trigonometric ratios of the given velocity components.
If we make a graph of the velocity vectors in their respective velocities according to the given description we will have something similar to the attached graph.
The angle could be obtained from the components of the opposite leg and the adjacent leg so that
The opposite leg value (y) is 40cm / s and the adjacent leg (x) is 30cm / s
Therefore the final direction that does the first ball is 36.87°
