Question

You throw a ball upwards at 22 m/s. How high will it go?

Answer 1

Answer:

24.69 meters

Explanation:

sorry if it's not right.

Answer 2

answer:

$h=24.69m$

step-by-step explanation:

$eg=mgh \\ek=\frac{1}{2} mv^2$

eg= gravitational energy

ek= kinetic energy

now, since no mass is given of the ball, both equations on their own do nothing for us, except leave us scratching our heads wondering how to figure out the problem. but, since the question states, “and no air resistance,” we now know, according to the law of conservation of energy, that the energy of the two equations will equal each other because none of the energy has dissipated or left the system.

the amount of energy present during the initial phase of the woman about to throw the ball will be present in the final phase, which will be at its highest point (according to this problem).

so now $eg=ek$

knowing this, we can now set the equations equal

$eg=ek\\mgh=\frac{1}{2} mv^2$

the two m’s cancel out, making the mass of the ball insignificant and not influential; next, substitute the values we are given in the problem

$(22m/s),(9.8m/s^2)\\m(9.8m/s^2)h=\frac{1}{2} m(22m/s)^2\\(9.8 m/s^2)h=\frac{1}{2} (22m/s)^2\\(9.8m/s^2)h=1/2 (484m^2/s^2)\\(9.8m/s^2)h=1/2 (242m^2/s^2)\\\\h= (242m^2/s^2)/(9.8m/s^2)$

as you can see, all units that need to be canceled out are indeed canceled, leaving us with just m, meters, which is what height is measured in

therefore, $h=24.69m$