Answer:
y = 9.64 m
Explanation:
This exercise should be solved using kinematics in one dimension, let's write the equations for the two cases presented
The rock is released
y = y₀ + V₀₁ t₁ - ½ g t₁²
In this case the speed starts is zero
y = y₀ - ½ g t₁²
The rock is thrown up
y = y₀ + v₀² t₂ -½ g t₂²
The height that reaches the floor is zero
y₀ - ½ g t₁² = y₀ + v₀₂ t₂ - ½ g t₂²
We use the initial velocity with the equation
v₂² = v₀₂² - 2 g y
At the point of maximum height v₂ = 0
v₀₂ = √ (2 g 
)
g (-t₁² + t₂²) = 2 √ (2 g ) t₂²
g (- 4.15² + 6.30²) = 2 √ (2 2 g) 6.3
g (22.4675) = 25.2 √ g
g² = 2²5.2 / 22.4675 g
g = 1.12 m / s²
Having the value of g we can use any equation to find the height
y = ½ g t₁²
y = ½ 1.12 4.15²
y = 9.64 m
Answer:
Vf = - 20 m/s ( -ve sign shows that the particle is moving opposite to positive x- direction).
Explanation:
Given:
Vi = 20 m/s, m= 10 mg =1 × 10⁻⁵ kg, q= -4.0 × 10⁻⁶ C , E= 20 N/C. t= 5.0 s
first to find Electric Force
F= Eq = 20 × -4.0 10⁻⁶ C = - 8 × 10⁻⁵ N (-ve sign shows that the field will push the particle opposite to positive x- direction)
We also have F=ma
⇒ a = F/m = - 8 × 10⁻⁵ N / 10 × 10⁻⁵ kg = -8 m/s² ( -ve sign shows that the particle is accelerated opposite to positive x- direction)
Now according the first equation of Motion.
Vf = Vi + at
Vf = 20 m/s + -8 m/s² × 5 s
Vf= -20 m/s ( -ve sign shows that the particle is moving opposite to positive x- direction)
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Answer:
Therefore, highest point that the cannon ball reaches is 168.7m
Explanation:
the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s
highest point that the cannon ball reaches?
g = 9.8m/s2
Therefore, highest point that the cannon ball reaches is 168.7m
