Answer:
The muzzle speed u =200 m/s
Explanation:
Let u be the muzzle speed.
maximum height s= 500 m
Assuming angle of elevation to be 30°
then initial vertical speed = usin30° = u/2
moreover, when the bullet reaches the maximum height its vertical velocity will be zero.
then using
[tex]v^2=u^2-2as[/tex]
a= 10 m/s^2
u= u/2
v= 0
s=500
we get
[tex]0^2=u^2/4-2\times10\times500[/tex]
u= 200 m/s.
Therefore, muzzle speed = 200 m/s
Answer:
The muzzle speed is 197.9 m/s.
Explanation:
Given that,
Maximum height = 500 m
Suppose the angle is 30°.
We need to calculate the muzzle speed
Using formula of maximum height
[tex]h_{max}=\dfrac{u^2}{2g}[/tex]
Where, h = maximum height
g = acceleration due to gravity
u = initial vertical velocity
Put the value into the formula
[tex]500=\dfrac{u^2\sin^{2}30}{2\times9.8}[/tex]
[tex]u^2=8\times500\times9.8[/tex]
[tex]u=\sqrt{4\times500\times9.8}[/tex]
[tex]u=197.9\ m/s[/tex]
Hence, The muzzle speed is 197.9 m/s.
