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A projectile thrown from the ground reached a horizontal displacement of 50 [m] and a maximum height of 100 [m]. What are the magnitude of the initial velocity v0 and the launching angle θ (withrespect to the horizontal) of the projectile? Use g = 9.81 m/s^2

User Mike Weir
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1 Answer

3 votes

Answer:


\theta=82.87^0

u = 44.44 m/s

Step-by-step explanation:

given,

horizontal displacement = 50 m

maximum height = 100 m

initial velocity (v₀) = ?

launching angle(θ) = ?

using formula


R = (u^2sin2\theta)/(g)........(1)


h = (u^2sin\theta)/(2g).........(2)

dividing equation (2)/(1)


(h)/(R) = ((u^2sin\theta)/(2g))/((u^2sin2\theta)/(g))


(h)/(R) =(sin^2\theta)/(2sin2\theta)


(4h)/(R) =tan \theta


\theta= tan{-1}{(4* 100)/(50)}


\theta=82.87^0

now using equation (2)


100 = (u^2sin82.87^0)/(2* 9.81)

u = 44.44 m/s

User Sukhpreet
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8.1k points