Answer:
1)
[tex]v_{oy}=11.29\ m/s[/tex]
2)
[tex]y=7.39\ m[/tex]
Explanation:
Projectile Motion
When an object is launched near the Earth's surface forming an angle [tex]\theta[/tex] with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
[tex]\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}[/tex]
Where [tex]y_o[/tex] is the initial height above the ground level, [tex]v_{oy}[/tex] is the vertical component of the initial velocity and t is the time
The y-component of the speed is
[tex]v_y=v_{oy}-gt[/tex]
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of [tex]v_o[/tex]
The object will reach the maximum height when [tex]v_y=0[/tex]. It allows us to compute the time to reach that point
[tex]v_{oy}-gt_m=0[/tex]
Solving for [tex]t_m[/tex]
[tex]\displaystyle t_m=\frac{v_{oy}}{g}[/tex]
Thus, the maximum heigh is
[tex]\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}[/tex]
We know this value is 8 meters
[tex]\displaystyle y_o+\frac{v_{oy}^2}{2g}=8[/tex]
Solving for [tex]v_{oy}[/tex]
[tex]\displaystyle v_{oy}=\sqrt{2g(8-y_o)}[/tex]
Replacing the known values
[tex]\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}[/tex]
[tex]\displaystyle v_{oy}=11.29\ m/s[/tex]
2) We know at t=1.505 sec the ball is above Julie's head, we can compute
[tex]\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}[/tex]
[tex]\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}[/tex]
[tex]\displaystyle y=1.5\ m+16,991\ m-11.098\ m[/tex]
[tex]y=7.39\ m[/tex]
