Answer:
Explanation:
Given
Initial Velocity (u)=13 m/s
angle[tex]=29^{\circ}[/tex]
distance between fence and deer=2.5 m
We consider deer jump similar to projectile motion
equation of trajectory
[tex]y=xtan\theta -\frac{gx^2}{2u^2(cos\theta )^2}[/tex]
[tex]y=2.5tan(29)-\frac{9.8\times 2.5^2}{2\times 13^2\times (cos29)^2}[/tex]
[tex]y=1.385-0.2368=1.148 m[/tex]
Thus deer will cross the fence with an difference in its jump and fence
1.5-1.148=0.351 m
[tex]h_{max}=\frac{u^2sin^2\theta }{2g}[/tex]
[tex]h_{max}=\frac{13^2\times (sin29)^2}{2\times 9.8}[/tex]
[tex]h_{max}=2.02 m[/tex]
so deer rises during when it is over fence
