Question

A stone is catapulted at time t = 0, with an initial velocity of magnitude 24.0 m/s and at an angle of 46.0° above the horizontal. (Neglect air resistance.) Find its horizontal and vertical displacements from the catapult site at the following times after launch.

(a) 0.80 s
m (horizontal)
m (vertical)
(b) 1.80 s
m (horizontal)
m (vertical)
(c) 3.53 s
m (horizontal)
m (vertical)

Answer 1

here,

the initial velocity of the stone , u = 24 m/s

theta = 46 degree

ux = 24 * cos(46)

ux = 16.67 m/s

uy = 24 * sin(46)

uy = 17.26 m/s

(a)

at t = 0.8 s

the horizontal displacement be x and vertical be y

x = ux * t

x = 16.67 * 0.8

x = 13.34 m

y = uy * t - 0.5 * g * t^2

y = 17.26 * 0.8 - 0.5 * 9.8 * 0.8^2

y = 10.67 m

the vertical and horizontal displacements are 10.67 m and 13.34 m

(b)

at t = 1.8 s

the horizontal displacement be x and vertical be y

x = ux * t

x = 16.67 * 1.8

x = 30.01 m

y = uy * t - 0.5 * g * t^2

y = 17.26 * 1.8 - 0.5 * 9.8 * 1.8^2

y = 15.19 m

the vertical and horizontal displacements are 1.38 m and 30.01 m

(c)

at t = 3.53 s

the horizontal displacement be x and vertical be y

x = ux * t

x = 16.67 * 3.53

x = 58.85 m

y = uy * t - 0.5 * g * t^2

y = 17.26 * 3.53 - 0.5 * 9.8 * 3.53^2

y = - 0.13 m

the vertical and horizontal displacements are 0.13 m below the level of catapult and 58.85 m