Answer: A projectile is any object in which the only force is gravity
Explanation: Equations on how to calculate projectile velocity is stated below:
The initial velocity Vo being a vector quantity, has two componentsVox and Voy
V0x = V0 cos(θ)
V0y = V0 sin(θ)
The acceleration A is a also a vector with two components Axand Ay given
Ax = 0 and Ay = - g = - 9.8 m/s2
Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant
Vx = Vocos(θ)
Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g
Vy = Vo sin(θ) - g t
Along the x axis the velocity Vx is constant and therefore the component x of the displacement is
x = Vocos(θ) t
Along the y axis, the motion is of uniform acceleration and the y component of the displacement is
y = Vo sin(θ) t - (1/2) g t2
Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F
| = I(
× 
)
given that
I = 2.6 A
= 0.17
= 0.52
so we substitute
|F | = 2.6( 0.17i" × 0.52j" )
|F | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)
For the bottom of window (position B)
We also have
Solving
So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at
Speed at which the ball passes the window’s top = 10.89 m/s
