A stone is vertically thrown upwards with an initial velocity of 40 m/s^1 calculate maximum height reached.
Answers
Step-by-step explanation:
A stone is thrown verticaly upward with an initial velocity of 40m/s. Taking g=10m/s2, find the maximum height reached by the stone. ... Total distance coverd by the stone =h+h=2h=2×80m=160m
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Answer:
The maximum height in meters reached by the stone thrown straight up with a speed of 35 m/s is 62.5 m as solved by my computer program. The answer is inside the box labelled as dymax. Vo = 35 m/s; the angle theta is 90 degrees and the value of the acceleration due to gravity g is 9.8m/s^2.
As the gravity always acts in downward direction the object thrown upwards will experience negative acceleration therefore the stone thrown vertically upwards its velocity is continuously decreased.
Then, explain your reasoning. At a projectile's highest point, its velocity is zero. At a projectile's highest point, its acceleration is zero.
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