Answer :
Answer:1.225 m
Explanation:
Given
Horizontal velocity before leaving ground [tex]=6\ m/s[/tex]
Range of jumper [tex]R=6\ m[/tex]
Suppose u is the velocity before leaving
[tex]u\cos \theta =6\ m/s[/tex]
Range of Projectile [tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]R=\frac{2(u\cos \theta )(u\sin \theta )}{g}[/tex]
[tex]6=\frac{2\times 6\times (u\sin \theta )}{g}[/tex]
[tex]u\sin \theta =4.9\ m[/tex]
Maximum height is given by
[tex]H=\frac{(u\sin \theta )^2}{2g}[/tex]
[tex]H=\frac{(4.9)^2}{2\times 9.8}[/tex]
[tex]H=1.225\ m[/tex]