Question

a body falls freely from rest.find (a)its accleration (b) the distance it falls in 3s (c) its speed after falling 70m (d) the time requied to reach a speed of 25 m/sec (e) the time taken to fall 300m.​

Answer 1

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]

Step-by-step 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.

`y = y_(0) +v_(0) *t+(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)+(1)/(2) *(9.81)*(3)^(2) \\\\y=44.14[m]`

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.

`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]`

`v_(f) ^(2) =0+2*9.81*(70)\\v_(f)=√(1373.4) \\v_(f)=37.05[m/s]`

d)

Using the following equation we can find the time for a 25 [m/s] speed.

`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]`

`25=0+(9.81*t)\\t=2.55[s]`

e)

The time taken to fall 300[m], can be found using the following equation:

`y=y_(0) +v_(0) *t+(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+(1)/(2) *(9.81)*t^(2) \\t=\sqrt{(2*300)/(9.81) } \\t=7.82[sg]`