Respuesta :
Answer:
a)Initial speed of the projectile = 196.2 m/s
b)Maximum altitude = 490.5 m
c) Range of projectile = 3398.28 m
d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Explanation:
Time of flight of a projectile is given by the expression,
[tex]t=\frac{2usin\theta}{g}[/tex]
Here θ = 30° and t = 20 s
a) [tex]t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s[/tex]
Initial speed of the projectile = 196.2 m/s
b) Maximum altitude is given by
[tex]H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m[/tex]
Maximum altitude = 490.5 m
c) Range of projectile is given by
[tex]R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m[/tex]
Range of projectile = 3398.28 m
d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s
Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s
We have equation of motion s = ut + 0.5 at²
Horizontal motion
u = 169.91 m/s
a = 0 m/s²
t = 15 s
Substituting
s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m
Vertical motion
u = 98.1 m/s
a = -9.81 m/s²
t = 15 s
Substituting
s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m
[tex]\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m[/tex]
Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
