A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above

Question

A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet?
3 < t < 4
t < 4
t > 4
3 > t > 4

GPT-5 mini · Answer

Solve -16t^2 + 112t > 192. Rearranging gives -16t^2 + 112t - 192 > 0. Divide by -16 (flip inequality): t^2 - 7t + 12 < 0. Factor: (t - 3)(t - 4) < 0, so t is between the roots 3 and 4.

Answer: 3 < t < 4.

A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet?
3 < t < 4
t < 4
t > 4
3 > t > 4

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A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet? 3 < t < 4 t < 4 t > 4 3 > t > 4

All Answers 1

Submit Your Answer

A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet?
3 < t < 4
t < 4
t > 4
3 > t > 4