You stick your arm over the edge of the Grand Canyon, and throw a ball vertically upwards with a velocity of 15 m/s. Compared to the heighht that you throw the ball, what height is the ball after 6 seconds?
t(up) = (Vf - Vo) / g,
t(up) = (0 - 15) / -9.8 = 1.53s. to reach max. ht.
hmax = (Vf^2 - Vo^2) / 2g,
hmax = (0 - (15)^2) / -19.6 = 11.5m.
above launching point.
h = Vo*t + 0.5g*t^2,
h = 15*6 - 4.9*6^2 = -86.4m = The distance below max. ht.
Free-fall distance=-86.4 - 11.5 = -98m
To determine the height of the ball after 6 seconds, we need to calculate its vertical displacement.
First, we can use the formula for vertical displacement of an object in free fall:
y = y0 + v0*t + (1/2)*a*t^2
Where:
- y is the final vertical position of the ball
- y0 is the initial vertical position of the ball (in this case, the height that you throw the ball from)
- v0 is the initial vertical velocity of the ball (15 m/s upwards in this case)
- a is the acceleration due to gravity (-9.8 m/s^2 for objects near the surface of the Earth)
- t is the time (6 seconds in this case)
Since you are throwing the ball upwards, the initial velocity (v0) will be positive, while the acceleration due to gravity (a) will be negative. We can substitute the values into the formula:
y = 0 + 15*6 + (1/2)*(-9.8)*(6^2)
Simplifying this equation gives us:
y = 0 + 90 + (-294)
Therefore, the height of the ball after 6 seconds is approximately -204 meters. The negative sign indicates that the ball is below the initial height.
Please note that this calculation is an idealized scenario and does not take into account air resistance or other factors that may affect the ball's motion.