Question

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

Answer 1

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