The height (in feet) of an object fired straight up from the ground level with a

Question

The height (in feet) of an object fired straight up from the ground level with an initial velocity of 200 feet per second is given by H(t)=-16t^2 +200t where t is the time (in seconds) A. How fast is the object moving after 1 second? B. At what is the object reaching the maximal height? C. How fast is the object when it hits the ground? D. During which interval of time is the speed decreasing? E. During which interval of time is the speed increasing?

Explanation / Answer

v = d H(t) / dt = 200 - 32t v = u + at So, a = -32 m/sec^2 a.) v = 200 - 32 = 168 m/sec b.) Time at which v becomes 0 => t = 200/32 sec = 25/4 sec c.) Hit the ground => H(t) = 0 => t = 200/16 = 25/2 sec d.) Speed decreases till it reaches maximum height => t = [0,25/4] sec e.) Speed decreases (negative velocity increases) after reaching maximum height t = [25/4,25/2] sec

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