Answer:
If max height = 1.1 meters, then initial velocity is 3.28 m/s
If max height is 1.1 feet, then the initial velocity is 5.93 ft/s
Explanation:
Recall the formulas for vertical motion under the acceleration of gravity;
for the vertical velocity of the object we have
[tex]v=v_0-g \,t[/tex]
for the object's vertical displacement we have
[tex]y-y_0=v_0\,t - \frac{g}{2} \,t^2[/tex]
If the maximum height reached by the object is given in meters, we use the value for g in [tex]m/s^2[/tex] which is: [tex]9.8\,\,m/s^2[/tex]
If the maximum height of the object is given in feet, we use the value for g in [tex]ft/s^2[/tex] which is : [tex]32\,\,ft/s^2[/tex]
Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:
[tex]v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}[/tex]
and now we use this to express the maximum height in the second equation we typed:
[tex]y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}[/tex]
Then if the max height is 1.1 meters, we use the following formula to solve for [tex]v_0[/tex]:
[tex]1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s[/tex]
If the max height is 1.1 feet, we use the following formula to solve for [tex]v_0[/tex]:
[tex]1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s[/tex]
