Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t = 
t = 
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
Answer:A) WHICH is 0kgm/s
Explanation:
Answer:
0.6kg
Explanation:
the unknown here is the mass of the second block
applying the law of the conservation of momentum
m₁v₁ + m₂v₂ = (m₁ + m₂) v₃
where m₁=mass of first block=2.2kg
m₂=mass of colliding block= ?
v₁= velocity of first block=1.2m/s
v₂=velocity of colliding block=4.0m/s
v₃= final velocity of combined block=1.8m/s
applying the formula above
(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8
2.64 + 4m₂ = 3.96 + 1.8m₂
collecting like terms
4m₂ - 1.8m₂ = 3.96 - 2.64
2.2m₂=1.32
divide both sides by 2.2
m₂= 0.6kg
Answer:
a) 14.2sec
b) 1394m away if horizontal speed never changes
c) 9.8m/s
Explanation:
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
Also, at any instant t
differentiate w.r.t.
