Aprojectile 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

u = 44.44 m/s

Explanation:

given,

horizontal displacement = 50 m

maximum height = 100 m

initial velocity (v₀) = ?

launching angle(θ) = ?

using formula

........(1)

.........(2)

dividing equation (2)/(1)

now using equation (2)

u = 44.44 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s