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:
14.2 m
Explanation:
Using conservation of energy:
PE at top = KE at bottom
mgh = ½ mv²
h = v² / (2g)
h = (16.7 m/s)² / (2 × 9.8 m/s²)
h = 14.2 m
Using kinematics:
Given:
v₀ = 16.7 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 14.2 m
Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer:
α = 3×10^-5 K^-1
Explanation:
let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.
then, the coefficient of linear expansion is given by:
α = ΔL/(ΔT×L)
= (0.3×10^-3)/(100)(100×10^-3)
= 3×10^-5 K^-1
Therefore, the coefficient of linear expansion is 3×10^-5 K^-1
Answer:
13.1
Explanation:
that's what I'm gonna go with, but u can research more
