Question

Gravitational acceleration on the moon is one sixth of that on Earth. A ball released from rest above the surface falls from height d in a time of t = 1.1 seconds.a) write an expression for the final velocity vf of the ball when it impacts the surface of the moon assuming it is dropped from rest. This expression should be in terms of only, g (gravitational acceleration on Earth) and time, tb) calculate the final velocity, vf, numerically in m/sc) calculate the height, d (in meters) from which the ball was droppedd) from how high up would this ball need to be dropped on the earth, De (in meters), if it took the same time to reach the ground as it did on the moon?

Answer 1

Answer:

$v_f=\dfrac{g}{6}\times t_b\ m/s$

1.7985 m/s

0.989175 m

5.93505 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v_f=u+at\\\Rightarrow v_f=0+\dfrac{g}{6}\times t_b\\\Rightarrow v_f=\dfrac{g}{6}\times t_b\ m/s$

The expression is $v_f=\dfrac{g}{6}\times t_b\ m/s$

$v_f=\dfrac{9.81}{6}\times 1.1\\\Rightarrow v_f=1.7985\ m/s$

The final velocity is 1.7985 m/s

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow d=0\times t+\dfrac{1}{2}\times \dfrac{9.81}{6}\times 1.1^2\\\Rightarrow d=0.989175\ m$

The distance dropped from the Moon is 0.989175 m

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow d=0\times t+\dfrac{1}{2}\times 9.81\times 1.1^2\\\Rightarrow s=5.93505\ m$

The distance dropped from the Earth is 5.93505 m

Answer 2

a).  $v_f=\left(-\frac{g}{6} \right) t$

b).  $v_f = -1.8 \ m/s$

c).  $d = 0.989 \ m$

d).  $d_e=5.93 \ m$

Given :

Gravitational acceleration on the moon =   $\frac{1}{6}$  of the gravitational acceleration on earth (g)

Time, $t=1.1$   s

Initial velocity, $u=0$  (since the body is at rest)

To find :

a). expression for final velocity of the ball,  $v_f$

b). Magnitude of final velocity, $v_f$  (in m/s)

c). height, $d$ (in meters) from which the ball was dropped

d). From how high up would the ball needs to be dropped on the earth, $d_e$ (in meters).

Solution :

a). Expression for final velocity of the ball in terms of 'g' and 't'

Using the formula of equation of motion :

$v=u+gt$

where, $v$  is final velocity

           $u$  is initial velocity

           $g$  is acceleration due to gravity

           $t$  is time

   

$\therefore$    $v=u+gt$  

     $v=0+\left( \frac{-g}{6} \right) \times t$

     $v=\left(-\frac{g}{6} \right) t$

or  $v_f=v=\left(-\frac{g}{6} \right) t$

Therefore, the expression of final velocity of the ball in terms of 'g' and 't'  is :

$v_f=v=\left(-\frac{g}{6} \right) t$

b).  Magnitude of final velocity, $v_f$

Using the same formula as above, we get :

$v_f=\left(-\frac{g}{6} \right) t$

$v_f=\left(-\frac{9.81}{6} \right) \times 1.1$

$v_f = -1.8 \ m/s$

Thus the final velocity is -1.8 m/s

c). Height, $h$ (in meters) from which the ball was dropped

Using the equation of motion

$S=ut+\frac{1}{2}gt^2$

where, $S$ = height

$\therefore \ \ \ \ S=ut+\frac{1}{2}gt^2$

       $S=(0)(1.1)+\frac{1}{2}\times \left(-\frac{9.81}{6} \right) \times (1.1)^2$

       $S=0.989 \ m$    (since, distance is positive only)

or    $d = S=0.989 \ m$

Therefore, the ball was dropped from a height of d = 0.989 m.

d). From what height the ball needs to be dropped on earth, $d_e$

Using the formula of equation of motion :

$S=ut+\frac{1}{2}at^2$

$S=(0)(1.1)+\frac{1}{2}(-9.81)(1.1)^2$

$S=5.93 \ m$  (since, distance is positive only)

or  $d_e=S=5.93 \ m$  

Thus,  the ball should be dropped from a height of  $d_e=5.93 \ m$  on earth if it takes the same time to reach the ground as it did on the moon.

Therefore :

a).  $v_f=\left(-\frac{g}{6} \right) t$

b).  $v_f = -1.8 \ m/s$

c).  $d = 0.989 \ m$

d).  $d_e=5.93 \ m$

Find out more about "object falling from height" here :

https://brainly.com/question/14580488