A baseball is thrown straight up at 15 m/s. How high will it go?

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A baseball is thrown straight up at 15 m/s. How high will it go?

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Answer 1 · 2016-04-16T20:35:38

I found $11.5 m$ $11.5m$

Explanation:

We can use here the general relationship from kinematics:

$v_{f}^{2} = v_{i}^{2} + 2 a (y_{f} - y_{i})$ $color(red)(v_f^2=v_i^2+2a(y_f-y_i))$

where:
$v_{i}$ $v_i$ is the initial velocity $= 15 \frac{m}{s}$ $=15m/s$ ;
$v_{f}$ $v_f$ is the final felocity which is zero in our case;
$a$ $a$ is the acceleraton of gravity $g = - 9.8 \frac{m}{s^{2}}$ $g=-9.8m/s^2$ (downward);
$y_{f}$ $y_f$ is the height reached from the ground where $y_{i} = 0$ $y_i=0$ .

So we get:

$0^{2} = 15^{2} - 2 \cdot 9.8 \cdot (y_{f} - 0)$ $0^2=15^2-2*9.8*(y_f-0)$

and:

$y_{f} = \frac{225}{19.6} = 11.5 m$ $y_f=(225)/(19.6)=11.5m$

Answer 2 · 2016-04-16T20:58:46

$h_{m} = 11, 47 m$ $h_m=11,47 "m"$

Explanation:

enter image source here
$h_{m} = maximum height$ $h_m="maximum height"$
$g = 9, 81 \frac{N}{k g}$ $g=9,81 N/(kg)$
$v_{i} = 15 \frac{m}{s} initial velocity$ $v_i=15 m/s" initial velocity"$
$α = \frac{π}{2}$ $alpha=pi/2$
$sin (\frac{π}{2}) = 1$ $sin (pi/2)=1$
$h_{m} = \frac{v_{i}^{2}}{2 \cdot g}$ $h_m=v_i^2/(2*g)$

$h_{m} = \frac{15^{2}}{2 \cdot 9, 81}$ $h_m=15^2/(2*9,81)$

$h_{m} = \frac{225}{19, 62}$ $h_m=225/(19,62)$

$h_{m} = 11, 47 m$ $h_m=11,47 "m"$

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A baseball is thrown straight up at 15 m/s. How high will it go?

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