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Question

From the edge of a cliff 48 m above the ground, a ball is vertically thrown upward at a velocity of 16 m/s. The function h(t)=−4t2+16t+48h(t)=$$-4t^2+16t+48  $$represents its height above the ground after it is thrown, where t is time in seconds. Determine the time at which the ball hits the ground.

Accepted Answer

t = 6 seconds

Answer

t = 2 seconds

Answer

t = 8 seconds

Answer

t = 4 seconds

From the edge of a cliff 48 m above the ground, a ball is vertically thrown upward at a velocity of 16 m/s. The function h(t)=−4t2+16t+48h(t)=-4t^2+16t+48 represents its height above the ground after it is thrown, where t is time in seconds. Determine the time at which the ball hits the ground.

From the edge of a cliff 48 m above the ground, a ball is vertically thrown upward at a velocity of 16 m/s. The function $h (t) = - 4 t^{2} + 16 t + 48$ represents its height above the ground after it is thrown, where t is time in seconds. Determine the time at which the ball hits the ground.