Explanation:
(a) What is the maximum height the arrow will attain?
Given:
v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s
vᵧ = 0 m/s
aᵧ = -9.8 m/s²
Find: Δy
vᵧ² = v₀ᵧ² + 2aᵧΔy
(0 m/s)² = (24.5 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 30.7 m
(b) The target is at the height from which the arrow was shot. How far away is it?
Given:
Δy = 0 m
v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s
aᵧ = -9.8 m/s²
Find: t
Δy = v₀ᵧ t + ½ aᵧt²
0 m = (24.5 m/s) t + ½ (-9.8 m/s²) t²
t = 5.00 s
Given:
v₀ₓ = 54 cos 27.0° m/s = 48.1 m/s
aₓ = 0 m/s²
t = 5.00 s
Find: Δx
Δx = v₀ₓ t + ½ aₓt²
Δx = (48.1 m/s) (5.00 s) + ½ (0 m/s²) (5.00 s)²
Δx = 241 m
Answer:
The time he can wait to pull the cord is 41.3 s
Explanation:
The equation for the height of the skydiver at a time "t" is as follows:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.
When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 m = 15000 m - 4.9 m/s² · t²
-15000 m / -4.9 m/s² = t²
t = 55.3 s
Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.
Answer:
The refractive index of glass, 
Solution:
Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.
To determine the refractive index of glass:
(1)
where
= refractive index of glass
= refractive index of glass
Now, using eqn (1)
