a projectile is fired straigt upward with a velocity of 400m/s its attitude above the ground t seconds after bieng fired is given by s(t)=-16t²+400t find the time after which the projectile hits the ground

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We are given the function for the altitude of a projectile as a function of time: s(t)=−16t2+400t, where t is the time in seconds and s(t) is the altitude in meters.To find the time after which the projectile hits the ground, we need to find the value of t when s(t)=0. In other words, we need to solve the equation:−16t2+400t=0We can factor out −16t to get:−16t(t−25)=0So the two solutions to this equation are t=0 and t=25. The solution t=0 corresponds to the initial time when the projectile was fired, so we can discard this solution.Therefore, the time after which the projectile hits the ground is t=25 seconds.

a projectile is fired straigt upward with a

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