A particle is moving along a projectile path where the initial height is 96 feet with an initial speed of 16 feet per second. What is the maximum height of the particle?

Question

anonymous · Answer

If we assume g is 32 feet per square second, and we want to find the maxinun possible height the ball can reach, then its initial velocity must be directed straight up.  Let h be the vertical distance from the ball's initial position to its position at time t, in seconds.  Assuming that up is the direction of increasing y, we know that v(t) = v(0) - g*t.  The ball reaches maximum altitude when its velocity is zero, at timt t = v(0)/g.  We also know that h = v(o)*t - (1/2)*g*t^2, so we can substitute v(0)/g for t into this to give us h = (v(0))^2/(2*g) = 16^2/(2*32) = 4 ft higher than its initial 96 feet.  Hence, the ball's maximum altitude is one hundred feet.

dls · Answer

13.06+96 feet=>109.06 feet approximately,what's the answer?