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 launch…

Question

asked Jun 6, 2020 117k views

1 Answer

Sukhpreet · Answer 1 · 2020-06-10T10:14:06+0000

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