A ball is thrown directly upward from the ground with an initial velocity of 4.8 ft./sec. Represent the height of the ball from the ground t seconds after it was thrown upward using the model h(t) .(1 point)
Responses
h(t)=−12⋅32t2+4.8t+4.8
h left parenthesis t right parenthesis equals negative Start Fraction 1 over 2 End Fraction times 32 t squared plus 4.8 t plus 4.8
h(t)=−12⋅9.8t2+4.8t
h left parenthesis t right parenthesis equals negative Start Fraction 1 over 2 End Fraction times 9.8 t squared plus 4.8 t
h(t)=−12⋅4.8t2+4.8t
h left parenthesis t right parenthesis equals negative Start Fraction 1 over 2 End Fraction times 4.8 t squared plus 4.8 t
h(t)=−16t2+4.8t
h left parenthesis t right parenthesis equals negative 16 t squared plus 4.8 t
h(t)=−16t2+4.8t
The correct representation for the height of the ball from the ground t seconds after it was thrown upward is:
h(t) = -16t^2 + 4.8t
To represent the height of the ball from the ground t seconds after it was thrown upward, we need to use the kinematic equation for vertical motion, which is given by:
h(t) = -16t^2 + v0t + h0
In this equation, h(t) represents the height of the ball at time t, v0 represents the initial velocity of the ball (in this case, 4.8 ft./sec), and h0 represents the initial height of the ball (which is 0 since the ball is thrown from the ground).
So, the correct representation for the height of the ball from the ground t seconds after it was thrown upward is:
h(t) = -16t^2 + 4.8t