Respuesta :
Answer:
Explanation:
Given
height of wall=5.15 m
angle of launch[tex](\theta )=55^{\circ}[/tex]
Launch velocity(u)=52.4 m/s
Time of flight will be sum of time of flight of projectile+time to cover 5.15 m
Time of flight of arrow[tex]=\frac{2usin\theta }{g}[/tex]
[tex]t=\frac{2\times 52.4\times sin55}{9.81}=8.76 s[/tex]
Now time require to cover 5.15 m
Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.
[tex]v_y=-52.4sin55[/tex] at zero vertical displacement
Thus time required will be [tex]t_2[/tex]
[tex]5.15=52.4sin55\times t+\frac{gt^2}{2}[/tex]
[tex]4.9t^2+42.92t-5.15=0[/tex]
[tex]t=0.118 s i.e. t_2=0.118 s[/tex]
total time =[tex]t_1+t_2=8.76+0.118=8.878 s[/tex]
(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s[tex](R_2)[/tex]
[tex]R_1=\frac{u^2sin2\theta }{g}=\frac{52.4^2\times sin110}{9.8}=263.28 m[/tex]
[tex]R_2=52.4cos55 \times 0.118=3.546 m[/tex]
[tex]R=R_1+R_2[/tex]=263.28+3.546=266.82 m
