a projectile is shot straight up with an initialvelocity of 512 ft/sec. Its heig

Question

a projectile is shot straight up with an initialvelocity of 512 ft/sec. Its height above the ground after tseconds is given by s(t) = 512t - 16t2 . a) find the velocity and acceleration after t seconds. b) what is the maximum heights attained? c) when does the projectile hit the ground and with whatvelocity. thanks again a projectile is shot straight up with an initialvelocity of 512 ft/sec. Its height above the ground after tseconds is given by s(t) = 512t - 16t2 . a) find the velocity and acceleration after t seconds. b) what is the maximum heights attained? c) when does the projectile hit the ground and with whatvelocity. thanks again

Explanation / Answer

Note that s(t) is the position function measuring the height of theprojectile in feet. The velocity function, v(t) is the first derivative of the positionfunction, that is v(t) = s'(t) The acceleration function, a(t) is the second derivative of theposition function, that is a(t) = s''(t) = v'(t) a) v(t) = 512 - 32t     a(t) = -32

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