To answer this question, consider the equation of motion of the ball:

`y_f = y_i + V_ot - (g*t^2)/2`

Here, `y_f` is the final height of the ball, `y_i` is the initial height of the ball, `V_o = 14 m/s` is the initial velocity directed upward, g = 9.8 m/s^2 is the acceleration due to gravity directed downward (hence, it is included with a negative sign), and t is the time.

We do not know the initial and final height of the ball, but we do know that it travels for 5 meters, so `y_f - y_i = d = 5 m` .

Plugging all the values in the equation, we can obtain a quadratic equation for t:

`5 = 14t - 9.8/2 t^2`

Simplifying this equation so that the leading coefficient is positive results in

`4.9t^2 - 14t +5 = 0`

We can use the quadratic formula to solve for t:

`t = (14 +- sqrt(14^2 - 4*4.9*5))/(2*4.9)`

As any quadratic equation, it has two roots: t = 2.44 s and t = 0.42 s. When the ball is thrown upward, it will pass the height of 5 meters on its way up, then it will go back down and again pass the height of 5 meters on its way down. The time when it will reach 5 meters going upward is the smaller one: t = 0.42 s.

**It will take the ball 0.42 s to reach 5 meters going upward.**

Another way to answer this question is by considering the equation relating the change of velocity to the distance passed:

`v_f^2 - v_i ^2 = -2gd`

Again, the acceleration due to gravity is taken with the negative sign because it directed downward, opposite to the direction of travel. From here, the final velocity of the ball when it reaches 5 meters can be found:

`v_f^2 = 14^2 - 2*9.8*5 = 98 m^2/s^2`

The final velocity is `v_f =sqrt(98) m/s` . Now, the time can be found from the equation for velocities:

`v_f = v_i - g*t`

`sqrt(98) = 14 - 9.8*t`

`t = (14 - sqrt(98))/9.8 =0.42 ` s

Again, **the time it will take the ball to reach 5 meters is 0.42 seconds.**

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