Answer:
a) acceleration = 9.81[m/s^2], b) distance = 44.1[m], c) v = 37.05[m/s], d) t= 2.54 [s], e) t = 7.82[s]
Explanation:
a) If the object falls freely its acceleration is equal to the gravity acceleration
a = g = 9.81[m/s^2].
b)
Using the following kinematic equation we can find the distance.
[tex]y = y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y_{0} =initial position =0[m]\\v_{0}=initial velocity = 0[m/s]\\y= distance traveled [m]\\t=time = 3[s]\\y = 0 +0 *(3)+\frac{1}{2} *(9.81)*(3)^{2} \\\\y=44.14[m][/tex]
Note: The initial velocity is zero because the body was released without velocity.
c)
Using the following kinematic equation we can find the velocity, with the 70[m] distance.
[tex]v_{f} ^{2} = v_{0} ^{2} +2*g*y\\where:\\v_{f}= final velocity [m/s]\\v_{0} = initial velocity [m/s]\\g = gravity [m/s]\\y=distance traveled[m][/tex]
[tex]v_{f} ^{2} =0+2*9.81*(70)\\v_{f}=\sqrt{1373.4} \\v_{f}=37.05[m/s][/tex]
d)
Using the following equation we can find the time for a 25 [m/s] speed.
[tex]v_{f} =v_{0}+(g*t )\\v_{0}=0[m/s]\\g=9.81[m/s^2]\\t = time[s]\\v_{f} =25[m/s][/tex]
[tex]25=0+(9.81*t)\\t=2.55[s][/tex]
e)
The time taken to fall 300[m], can be found using the following equation:
[tex]y=y_{0} +v_{0} *t+\frac{1}{2} * g*t^{2} \\where:\\y = distance traveled=300[m]\\v_{0}=0[m/s]\\g=9.81[m/s^2]\\300=(0)*t+\frac{1}{2} *(9.81)*t^{2} \\t=\sqrt{\frac{2*300}{9.81} } \\t=7.82[sg][/tex]
