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
<u>Vertical position of the rock at t = 2 seconds</u>
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
<u>Horizontal position of the rock at t = 2 seconds</u>
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
