Question

Your favorite professor decided to throw in the air a marker above his head (vertically). If the marker leaves his hand at a height of 1.2 m from the ground with an initial velocity of 20 m/s, when would the marker reach the highest point in the air (1.6 m from the ground)?

Answer 1

Answer:

$t=4.06s$

Explanation:

From the exercise we know

$y_{o}=1.2m\\v_{o}=20m/s\\y=1.6m\\g=-9.8m/s^2$

To find how long does it takes the marker to reach the highest point we need to use the equation of position:

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$1.6m=1.2m+(20m/s)t-\frac{1}{2}(9.8m/s^2)t^2$

$0=-0.4+20t-4.9t^2$

Now, we need to use the quadratic formula:

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=-0.4$

Solving for t

$t=0.020s$ or $t=4.06s$

So, the answer is t=4.06s because the other option is almost 0 and doesn't make any sense for the motion of the marker