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Aloiza [94]
3 years ago
8

A projectile is thrown upward so that its distance above the ground after t seconds is h=-121t^2 +504t. After how many seconds d

oes it reach it maximum height?
Mathematics
1 answer:
lora16 [44]3 years ago
6 0

Answer:

<em> 2.08secs</em>

Step-by-step explanation:

Given the equation modeled by the height as h=-121t^2 +504t

The velocity of the object at the max height is zero, hence;

v = dh/dt = 0

dh/dt = -2(121)t + 504

dh/dt = -242t + 504

0 = -242t + 504

242t = 504

t = 504/242

t = 2.08secs

<em>Hence the projectile reaches its maximum height after 2.08secs</em>

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2 years ago
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