Answer:
Vi = 94.64 m/s
Explanation:
I order to find out the initial velocity of the object, we can use third equation of motion:
2ah = Vf² - Vi²
where,
a = acceleration = -9.8 m/s²
h = maximum height covered by object = 460 m - 3 m = 457 m
Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)
Vi = Initial Velocity = ?
Therefore,
2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²
Vi = √8957.2 m²/s²
<u>Vi = 94.64 m/s</u>
Your not really supposed to date your relative
The net force will point towards the acceleration of the object, as supported by Newton's second law.
Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
Answer:
1317.4 m
Explanation:
We are given that
Angle=
Initial speed =
We have to find the horizontal distance covered by the shell after 5.03 s.
Horizontal component of initial speed=
Vertical component of initial speed=
Time=t=5.03 s
Horizontal distance =
Using the formula
Horizontal distance=
Horizontal distance=1317.4 m
Hence, the horizontal distance covered by the shell=1317.4 m