We have the following equation for height: h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0 Where, a: acceleration vo: initial speed h0: initial height. The value of the acceleration is: a = -g = -9.8 m / s ^ 2 For t = 0 we have: h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0 h (0) = h0 h0 = 0 (reference system equal to zero when the ball is hit). For t = 5.8 we have: h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0 vo = (1/2) * (9.8) * (5.8) vo = 28.42 Substituting values we have: h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0 Rewriting: h (t) = -4.9 * t ^ 2 + 28.42 * t The maximum height occurs when: h '(t) = -9.8 * t + 28.42 -9.8 * t + 28.42 = 0 t = 28.42 / 9.8 t = 2.9 seconds. Answer: The ball was at maximum elevation when: t = 2.9 seconds.
Vertical Displacement is zero because the initial and final height above the ground is same.
Total time of flight is = 5.8 s
The time taken by ball to reach the maximum elevation is same as the time to fall back the same height. The time at which the ball was at maximum elevation is half the time of flight.
Thus, the time at which the ball was at maximum elevation =
8 miles per hour because if it is moving at 4 miles every half hour that means you have to multiply it by two to get it in miles per hour and we all know that 4 times 2 is 8 so it would be 8 miles per hour =)
the rock will continue at the same speed unless it is affected by another force such as gravity and so if you threw it it will continue to move unless affected by a force
Explanation:
this is because Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
