Respuesta :
Answer:
The force is calculated as 338.66 N
Explanation:
We know that force is given by
[tex]\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{m(v_{f}-v_{o})}{\Delta t}[/tex]
We know that range of a projectile is given by
[tex]R=\frac{v_{o}^{2}sin(2\theta )}{g}[/tex]
it is given that R=130 m applying values in the above equation we get
[tex]R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s[/tex]
Thus the force is obtained as
[tex]\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2\times 10^{}-3(36.04-0)}{6.3\times 10^{-3}}[/tex]
Thus force equals [tex]F=338.66N[/tex]
Answer:
522.84 N
Explanation:
m = 59.2 g = 59.2 x 10^-3 kg
θ = 50.5
R = 130 m
g = 9.8 m/s^2
t = 6.3 m s = 6.3 x 10^-3 s
Let u be the velocity of projection
Use the formula for the range
[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]
130 = u^2 x Sin 2(50.5) / 9.8
u = 36.06 m/s
The vector form of the initial velocity
u = 36.05 (Cos 50.5 i + Sin 50.5 j)
u = 22.93 i + 27.82 j
The velocity of the body as it strikes with the surface is same but the direction is downward.
v = 22.93 i - 27.82 j
Change in momentum, Δp = m (v - u)
Δp = 59.2 x 10^-3 x (22.93 i - 27.82 j - 22.93 i - 27.82 j)
Δp = 59.2 x 10^-3 x ( - 55.64 j)
Δp = - 3.294 j
Δp = 3.24
Force = Rate of change in momentum
F = Δp / Δt = 3.294 / (6.3 x 10^-3) = 522.84 N
