Question

if a stone is dropped from a height 122.5m.if it is stopped after 3 seconds and released again...what is the time of descent??​

Answer 1

Answer:

367.5

Explanation:

122.5 × 3 is 367.5

Answer 2

The height from which the stone is dropped = hAcceleration due to gravity experienced = gInitial velocity , u = 0To find :The time taken by the stone to reach on the ground .Solution :Let the time taken by the stone to reach the ground be t .By using 2nd equation motion ,     distance = initial velocity * time taken + (1/2) * acceleration * time * time=>  -h = u * t - (1/2) * g * t * t( minus sign for height and g as per convention )=>  -h = - (1/2) * g * t * t=>  t = is the time taken by the stone to reach on the groundWhere u is initial velocity, ‘s’ is distance travelled (Height of the tower in this case), ‘t’ is time travelled and ‘a’ is the constant acceleration.So , here as the object is dropped , the initial velocity ,u is 0Acceleration is the acceleration due to gravity ‘g’ which is constant and has value of 9.8m/ s2 Time travelled is given as t=10 seconds

Substituting the values in the equation

s=ut+( 1/2) a t2 we gets=0 * 10 + ( 1/2) * 9.8 * 10 *10s=490 meters

So, the height of the tower is 490 meters

For this question we can use Newton's 2 equations=ut +1/2at^2.

Here h is s and g is 10(we generally consider gravity as 10)

u=initial velocity which is 0

Therefore

H = 5t^2t=√h/√5.the value of t is as Above

Thank you