Question

Question 4 (2 points)

Answer 1

Answer:

t = 1.15 + .713 = 1.863 [s]

Explanation:

To solve this problem we must use the following kinematic equation, but first we identify the initial data.

vo = 7 [m/s]

yo = 4 [m]

g = 9.81[m/s^2]

$v_{f}=v_{o}-g*t\\0 = 7 - (9.81*t)\\7 = 9.81*t\\t = 0.713[s]$

The final velocity happens at the moment when the maximum height is reached, at this point the final speed is zero.

$y=y_{o}+v_{o}*t-0.5*g*t^{2}\\ y = 0+ (7*0.713)-0.5*9.81*(0.713)^{2}\\ y = 2.5[m]$

The total elevation will be 2.5 + 4 = 6.5 [m]

Now using again the total height we can find the final velocity.

$v_{f}^2=v_{o}^2+2*g*y\\v_{f}=\sqrt{2*g*y}\\ v_{f}=\sqrt{2*9.81*6.5}\\ v_{f}=11.3[m/s]$

With this final velocity we cand find the time.

$v_{f}=v_{i}+g*t\\11.3=0+9.81*t\\11.3=9.81*t\\t=1.15[s]$

Now we have to sum the two times, the time it takes to go up and the time it takes to go down.

t = 1.15 + .713 = 1.863 [s]