If an object is thrown upward subject only to gravity, the height (in ft) of the projectile t seconds after launch can be modeled by h(t) = −16t2 + v0t + h0 Where h0 is the height of the projectile when t = 0 and v0 is the velocity (in ft/s) when t = 0 Assume a major league pitcher throws a baseball is thrown upward off a 500ft tall cliff overlooking water with a velocity of 132ft/s. Show work to receive credit When will the ball be even with the cliff on the way back down? (as in, when will the ball be 500ft above the water?) After how many seconds will the baseball hit the water below the cliff? (round to 2 decimal places)

Q: A ball is thrown so that the motion is defined by the equations x = 4t    and  y=2+6t−4.9t2  ⁢…

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Q: km E SE km/h W km/h W km/h W O km/h W km/h W km/h W km/h W km/h W

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Projectile Motion