Question

anders throws a baseball straight up from a height of 3 feet above the ground. the initial velocity of the baseball is 25 feet per second. what is the height of the ball 1.5 seconds after it is thrown? round your answer to the nearest hundredth. (1 point)
seconds

Explanation:

Step1: Recall the height formula for projectile motion

The formula for the height \( h \) of an object in vertical motion is \( h(t)=-16t^{2}+v_{0}t + h_{0} \), where \( v_{0} \) is the initial velocity, \( h_{0} \) is the initial height, and \( t \) is the time. Here, \( v_{0} = 25\) ft/s, \( h_{0}=3\) ft, and \( t = 1.5\) s.

Step2: Substitute the values into the formula

Substitute \( t = 1.5\), \( v_{0}=25\), and \( h_{0} = 3\) into the formula:
\[

$$\begin{align*} h(1.5)&=- 16\times(1.5)^{2}+25\times1.5 + 3\\ &=-16\times2.25+37.5 + 3\\ &=-36 + 37.5+3\\ &=1.5 + 3\\ &=4.5 \end{align*}$$

\]

Wait, there is a mistake in the calculation above. Let's recalculate:

\[

$$\begin{align*} h(1.5)&=-16\times(1.5)^{2}+25\times1.5 + 3\\ &=-16\times2.25+37.5 + 3\\ &=- 36+37.5 + 3\\ &=( - 36+37.5)+3\\ &=1.5 + 3\\ &=4.5? \text{ No, wait, }-16\times2.25=-36, 25\times1.5 = 37.5,-36 + 37.5=1.5,1.5 + 3 = 4.5? \text{ But let's check again. } \end{align*}$$

\]

Wait, no, the correct formula for the height of an object thrown vertically (under the influence of gravity near the surface of the Earth, where the acceleration due to gravity is \( 32\) ft/s², so the formula is \( h(t)=-16t^{2}+v_{0}t + h_{0} \) because the acceleration \( a=- 32\) ft/s², and using the kinematic equation \( h = h_{0}+v_{0}t+\frac{1}{2}at^{2}\), so \( \frac{1}{2}a=\frac{1}{2}\times(- 32)=-16\)).

Let's recalculate:

\( t = 1.5\), so \( t^{2}=2.25\)

\( - 16\times2.25=-36\)

\( 25\times1.5 = 37.5\)

\( h_{0}=3\)

So \( h(1.5)=-36 + 37.5+3=4.5\)? Wait, that seems low. Wait, maybe I made a mistake. Wait, initial velocity is 25 ft/s, initial height 3 ft. Let's check with \( t = 0\), \( h(0)=3\), which is correct. At \( t = 1\) second: \( h(1)=-16\times1+25\times1 + 3=-16 + 25+3 = 12\) ft. At \( t = 1.5\):

\( -16\times(1.5)^2=-16\times2.25=-36\)

\( 25\times1.5 = 37.5\)

\( -36+37.5 = 1.5\)

\( 1.5 + 3=4.5\). Wait, but that would mean the ball is at 4.5 feet at 1.5 seconds. But let's think about the time when it reaches the maximum height. The time to reach max height is \( t=\frac{v_{0}}{32}=\frac{25}{32}\approx0.78125\) seconds. So after 0.78 seconds, it starts to fall. So at \( t = 1.5\) seconds, which is after the max height, the height should be less than the max height. The max height is \( h(\frac{25}{32})=-16\times(\frac{25}{32})^2+25\times\frac{25}{32}+3\). Let's calculate that:

\( (\frac{25}{32})^2=\frac{625}{1024}\)

\( -16\times\frac{625}{1024}=-\frac{10000}{1024}\approx - 9.765625\)

\( 25\times\frac{25}{32}=\frac{625}{32}\approx19.53125\)

So max height \( h\approx - 9.765625+19.53125 + 3=12.765625\) ft. Then at \( t = 1.5\) seconds, which is \( 1.5 - 0.78125 = 0.71875\) seconds after reaching max height. The distance fallen in that time is \( \frac{1}{2}\times32\times(0.71875)^2=16\times0.5166015625 = 8.265625\) ft. So the height should be \( 12.765625-8.265625 = 4.5\) ft. So the calculation is correct.

Answer:

\( 4.50\) (rounded to the nearest hundredth)