a stone is thrown vertically upwards with an initial velocity of20 m/ sec.find the maximum height itreaches and the time taken by it to reach the height.( g= 10 m/ s2)
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Answer:
The speed of a rising body is being decelerated by the Earth’s force of gravity by a value equals to g. The speed is governed by this equation v = gt. Since v = 20 m/s and g is 10 m/s^2 downward, the time of rising will stop after 2 seconds. That is -10 m/s * 2 = -20 m/s downward. When a speed of 20 m/s upward is added to -20 m/s downward the net velocity is zero.
The time to reach its maximum height is 2 seconds. To solve for the maximum height, the formula is y = voy * t - 1/2 gt^2 where y is the vertical displacement upward, voy is the initial vertical velocity.
Plugging in the data to the equation y= 20m/s * 2s -1/2 * 10 m/s^2 * (2s)^2
y = 40 m - 1/2 * 10 * 4s^2
y = 40 m - 20 m
y = 20 m
The maximum height reached by the stone is 20 meters and the time taken to reach the maximum height is 2 seconds.
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