Answer:
The angle of launch is 52.49 Degree.
Explanation:
The Range R and Height H of a thrown object is calculated using the formula,
R=V₀² sin(2φ)/g
H=V₀²sin²(φ)/g
From these equations it can be written,
V₀²=R g/ sin(2φ)
V₀²=H g/ sin²(φ)
These values are equal so it can be written by equating these equations,
R g/sin(2φ)=H g/sin²(φ)
tan(φ)= 2H/R
Given H=72.3 m and R=111 m, the angle of launch is,
tan(φ)= 2*72.3/111
φ= 52.49 Degree.
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I think the anwser is 3)calculus
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
