Question

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.71 s, and for the (e) horizontal and (f) vertical components at t = 5.44 s. Assume that the catapult is positioned on a plain horizontal ground.

Answer 1

Part a)

`x = 15.76 m`

Part b)

`y = 7.94 m`

Part c)

`x = 26.16 m`

Part d)

`y = 7.49 m`

Part e)

`x = 83.23 m`

Part f)

`y = -75.6 m`

Step-by-step explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

`v_x = 19.9 cos39.9`

`v_x = 15.3 m/s`

similarly we have

`v_y = 19.9 sin39.9`

`v_y = 12.76 m/s`

Part a)

Horizontal displacement in 1.03 s

`x = v_x t`

`x = (15.3)(1.03)`

`x = 15.76 m`

Part b)

Vertical direction we have

`y = (12.76)(1.03) - 4.9(1.03)^2`

`y = 7.94 m`

Part c)

Horizontal displacement in 1.71 s

`x = v_x t`

`x = (15.3)(1.71)`

`x = 26.16 m`

Part d)

Vertical direction we have

`y = (12.76)(1.71) - 4.9(1.71)^2`

`y = 7.49 m`

Part e)

Horizontal displacement in 5.44 s

`x = v_x t`

`x = (15.3)(5.44)`

`x = 83.23 m`

Part f)

Vertical direction we have

`y = (12.76)(5.44) - 4.9(5.44)^2`

`y = -75.6 m`