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kaheart [24]
3 years ago
13

A projectile thrown from the ground reached a horizontal displacement of 50 [m] and a maximum height of 100 [m]. What are the ma

gnitude of the initial velocity v0 and the launching angle θ (withrespect to the horizontal) of the projectile? Use g = 9.81 m/s^2
Physics
1 answer:
Semenov [28]3 years ago
5 0

Answer:

\theta=82.87^0

u = 44.44 m/s

Explanation:

given,

horizontal displacement = 50 m

maximum height = 100 m

initial velocity (v₀) = ?

launching angle(θ) = ?

using formula

R = \dfrac{u^2sin2\theta}{g}........(1)

h = \dfrac{u^2sin\theta}{2g}.........(2)

dividing equation (2)/(1)

\dfrac{h}{R} = \dfrac{\dfrac{u^2sin\theta}{2g}}{\dfrac{u^2sin2\theta}{g}}

\dfrac{h}{R} =\dfrac{sin^2\theta}{2sin2\theta}

\dfrac{4h}{R} =tan \theta

\theta= tan{-1}{\dfrac{4\times 100}{50}}

\theta=82.87^0

now using equation (2)

100 = \dfrac{u^2sin82.87^0}{2\times 9.81}

u = 44.44 m/s

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Answer:

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Explanation:

From the question we are told that

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Given that the two car are now in the same position we have that

    119  + 14.64 *  t  =   0   +  (18.3 * t )

   t =  32.5 \  s

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