Question

Vertical Motion The height of an object moving vertically is given by $$s=-16 t^{2}+96 t+112$$ with \(s\) in \(\mathrm{ft}\) and \(t\) in sec. Find (a) the object's velocity when \(t=0,\) (b) its maximum height and when it occurs, and (c) its velocity when \(s=0\) .

Short answer

The object's velocity at \(t = 0\) is 96 ft/sec. It reaches a maximum height of 160 feet at \(t = 3\) seconds. When the object hits the ground \(s = 0\), its velocity is -128 ft/sec.

Step-by-step solution

Find the object's velocity at \(t = 0\)

The formula for the height of an object thrown vertically up in the air is given by \(s = -16t^2 + 96t + 112\). Here, the velocity of the object is represented by the derivative of the height function. Differentiate the height, \(s(t)\), with respect to time, \(t\), to find the velocity function, \(v(t)\).\n\nThe derivative of the function \(s(t) = -16t^2 + 96t + 112\) with respect to \(t\) is \(v(t) = -32t + 96\). Hence the object's velocity function is \(v(t) = -32t + 96\), where \(v\) is in ft/sec. To find the object's velocity at \(t = 0\), put \(t = 0\) in \(-32t + 96\) which gives \(v = 96\) ft/sec.

Calculate the maximum height and when it occurs

To find the maximum height and when it occurs, first set the velocity equation \(-32t + 96 = 0\) and solve for \(t\), after which we substitute this time value in the equation for \(s(t)\) to find the maximum height. Solving the equation for \(t\) gives \(t = 3\) seconds. Substitute \(t = 3\) into the height equation \(s(t) = -16(3)^2 + 96(3) + 112\) to find the maximum height. The maximum height is \(s = 160\) feet. Therefore, the object reaches a maximum height of 160 feet at \(t = 3\) seconds.

Find the velocity when \(s = 0\)

To find the velocity when the object hits the ground, \(s = 0\), we first find the time when the object hits the ground by solving the height equation \(-16t^2 + 96t + 112 = 0\) for \(t\). The resulting values of \(t\) are 0 and 7 seconds (ignoring the negative solution). We choose \(t = 7\) seconds because at \(t = 0\), the object was launched. Substituting \(t = 7\) into the velocity equation gives \(v(7) = -32(7) +96 = -128\) ft/sec. So, the object is travelling at a velocity of -128 ft/sec when it hits the ground.