Respuesta :
Answer:
Explanation:
Given
mass of projectile [tex]m=6.8\ kg[/tex]
initial horizontal speed [tex]u_x=14.5\ m/s[/tex]
height [tex]h=26.7\ m[/tex]
Considering vertical motion
velocity gained by projectile during 26.7 m motion
[tex]v^2-u^2=2 as[/tex]
v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex]v^2-(0)^2=2\times (9.8)\times (26.7)[/tex]
[tex]v=\sqrt{523.32}[/tex]
[tex]v=22.87\ m/s[/tex]
Horizontal velocity will remain same as there is no acceleration
final velocity [tex]v_{net}=\sqrt{(v)^2+(u_x)^2}[/tex]
[tex]v_{net}=\sqrt{733.57}=27.08\ m/s[/tex]
Initial kinetic Energy [tex]K_i=\frac{1}{2}mu_x^2[/tex]
[tex]K_i=\frac{1}{2}\times 6.8\times (14.5)^2=714.85\ J[/tex]
Final Kinetic Energy [tex]K_f=\frac{1}{2}mv_{net}^2[/tex]
[tex]K_f=\frac{1}{2}\times 6.8\times (27.08)^2[/tex]
[tex]K_f=2493.30\ J[/tex]
Work done by all the force is equal to change in kinetic Energy of object
Work done by gravity is [tex]W_g[/tex]
[tex]W_g=\Delta K[/tex]
[tex]W_g=2493.30-714.85=1778.45\ J[/tex]
