a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity
1 answer:
consider the motion along the X-direction
X = horizontal displacement = 80 m
= initial velocity along the x-direction = v Cos60
t = time of travel
using the equation
X = t
80 = (v Cos60) (t)
t = 160/v eq-1
consider the motion in vertical direction :
Y = vertical displacement = 20 m
= initial velocity in Y-direction = v Sin60
a = acceleration = - 9.8 m/s²
t = time of travel = 160/v
using the equation
Y = t + (0.5) a t²
20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²
v = 32.5 m/s
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Answer:
13.37 rev/min
Explanation:
acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration () = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².
r = 9 m
Centripetal acceleration () is given by:
The velocity (v) is given by:
v = ωr; where ω is the angular velocity
Hence:
ω = v/r = 12.6 / 9
ω = 1.4 rad/s
ω = 2πN
N = ω/2π = 1.4 / 2π
N = 0.2228 rev/s
N = 13.37 rev/min
Answer:
A) T.
Explanation:
Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:
So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.