After catching the ball, sarah throws it back to julie. the ball leaves sarah's hand a distance 1.5 meters above the ground. it is moving with a speed of 9 m/s when it reaches a maximum height of 7 m above the ground. what is the speed of the ball when it leaves sarah's hand?

vo = 13.74 m/s

Explanation:

The goal is to find the initial velocity vo. According to the exercise, the maximum height that the ball will reach and the initial height will be respectively, yo = 1.5 m and ymax = 7 m, tmax will be equal to the time it takes for the ball to reach the maximum height. Having the equation:

vx = vo * cos∝o

clearing vo:

vo = vx/cos∝o

vy = vo * sin∝o + gtmax

0 = sin∝o * (vx/cos∝o) + gtmax

0 = vx * tan∝o + gtmax

gtmax = -vx * tan∝o

clearing tmax:

tmax = - (vx * tanao/g)

ymax = i + (vo * sin∝o) * tmax + (gtmax^2/2)

Replacing:

ymax = i - (vx^2 * tan^2∝o/2 * g)

clearing the angle a or:

∝o = arctan (((2 * g * (i - ymax)) / vx^2)^1/2)

substituting values:

∝o = arctan (((2 * (- 9.8) * (1.5 - 7)) / (9^2))) = 49.08 °

We find the initial speed with the following formula:

vo = vx/cos∝o = 9/cos49.08 ° = 13.74 m/s