Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² = 
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo = ![v_{oy}² /2 g y- 0 = 10.0²/2 9.8 y - 0 = 5.10 m The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system y₂ = 5.1 + 44 y₂ = 49.1 m Let's use the other equation to find the time [tex]v_{y}](https://tex.z-dn.net/?f=v_%7Boy%7D%C2%B2%20%2F2%20g%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y-%200%20%3D%2010.0%C2%B2%2F2%209.8%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y%20-%200%20%3D%205.10%20m%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%0A%3C%2Fp%3E%3Cp%3EThe%20height%20from%20the%20ground%20is%20the%20height%20that%20rises%20from%20the%20reference%20system%20plus%20the%20depth%20of%20the%20ground%20from%20the%20reference%20system%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%205.1%20%2B%2044%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%2049.1%20m%0A%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ELet%27s%20use%20the%20other%20equation%20to%20find%20the%20time%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20%20%5Btex%5Dv_%7By%7D)
= - g t
t = / g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 = ² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
Answer:
0
Explanation:
Given the following :
Height of wall = 19.2 m
Time taken to hit the ground = 5 seconds
Acceleration due to gravity (g) = 9.8m/s ( downward motion)
The initial velocity of the object refers to the Velocity of the object at time t = 0
Initial Velocity = g × time
Initial Velocity = 9.8 × 0 = 0
E=kq/r^2
q=(E*r^2)/k
q=(.086N/C)(1.7m^2)/(8.99*10^9N*m^2/C^2)
q=2.76*10^-11 C
q=2.8*10^-11 C
Answer:
0.046
Explanation:
displacement = velocity/ time
d = 6m/s / 130s
d = 0.046m
