Question

Can someone please tell me how to solve these question below.

Answer 1

Answer:

Vertical position of the rock after 2 seconds = 11.46 m

Horizontal position of the rock after 2 seconds = 13.06 m

Explanation:

Initial height = 25 m

Angle of throw = 25 degrees

Velocity of throw = 7.2 m/s

Horizontal velocity of throw = 7.2 cos 25 = 6.53 m/s

Vertical velocity of throw = 7.2 sin 25 = 3.04 m/s

Acceleration in horizontal direction = 0 m/s²

Acceleration in vertical direction = -9.8 m/s²

Vertical position of the rock at t = 2 seconds

We have equation of motion s = ut + 0.5at²

                                     s = 3.04 x 2 - 0.5 x 9.81 x 2²

                                     s = -13.54 m

Vertical position = 25 - 13.54 = 11.46 m

Horizontal position of the rock at t = 2 seconds

We have equation of motion s = ut + 0.5at²

                                     s = 6.53 x 2 + 0.5 x 0 x 2²

                                     s = 13.06 m

Horizontal position = 13.06 m

Vertical position of the rock after 2 seconds = 11.46 m

Horizontal position of the rock after 2 seconds = 13.06 m