A projectile is launched at ground level with an initial speed of 49.5 m/s at an angle of 40.0° above the horizontal. It strikes a target above the ground 3.50 seconds later. What are the x and y distances from where the projectile was
launched to where it lands?
x distance
m
y distance
m

x = 132.7 m

y = 51.34 m

Explanation:

Given :

Initial speed, u = 49.5 m/s²

Angle of projection, θ = 40°

Time, t = 3.50 seconds

The distance, x = horizontal component ;

Distance = speed * time

Distance = uCosθ * 3.50

Distance = 49.5 * Cos40° * 3.50

Distance = 49.5 * Cos40° * 3.50

Horizontal distance = 132.7 m

Vertical distance, y :

Sy = ut + 1/2gt²

Sy = Vertical distance ; g = 9.8 m/s²

Sy = 49.5 * sin40 * 3.5 - (0.5 * 9.8 * 3.5²)

Sy = 111.36295 - 60.025

Sy = 51.33795 m

x = 132.7 m

y = 51.34 m

i think its

the motion of a pendulum is a classic example of mechanical energy conservation. a pendulum consists of a mass (known as a bob ) attached by a string to a pivot point. as the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion.