Answer:
1.95m
Explanation:
To calculate the vertical velocity needed we use the formula [tex]v^2=v_0^2+2ad[/tex], where we will take the upwards direction as the positive one (gravity will be negative). We will use it only for the vertical component, and we know that at maximum height [tex]d_y[/tex] the final vertical velocity must be 0m/s, so the initial velocity to achieve that displacement must be:
[tex]v_{y0}=\sqrt{v_y^2-2a_yd_y}=\sqrt{(0m/s)^2-2(-9.8m/s^2)(0.75m)}=3.83m/s[/tex]
How far in the horizontal direction must be calculated with the formula [tex]d_x=v_x t[/tex] since there is no acceleration in that component, but we need the time he needed to do it. We will obtain it using the equation[tex]v=v_0+at[/tex], and looking at the vertical component again (since time will be the same for both components, something obvious but powerful):
[tex]t=\frac{v_y-v_{0y}}{a_y}[/tex]
We can substitute this on the previous equation:
[tex]d_x=v_x \frac{v_y-v_{0y}}{a_y}[/tex]
And using our values we get:
[tex]d_x=(5m) \frac{(0m/s)-(3.83m/s)}{(-9.8m/s)}=1.95m[/tex]
