Question

An object falls from a bridge that was 45m
above the water. In water a small row boat
was moving with velocity u and was at a
distance 13.5m from the point of impact when object
was released. At the instant when the object
released if the boat start accelerating at 1 m/s2,
so that if falls directly into the small row
boat then in the initial velocity u' of boat
(g=10m/s)​

Answer 1

Answer:

The boat was moving with an initial velocity u' of 3m/s

Explanation:

Let The initial velocity of the boat be u'

Now let stone fall in time t sec with acceleration of 10m/s^2

So by Second equation of motion we get

Height of bridge = 45

S = u*t + 1/2 g*t^2

45 = 0*t + 1/2 10*t^2

45 = 5t^2

9 = t^2

So, t = 3sec

Now,

Boat  is 13.5 m away  s= 13.5m

a = 1m/s^2

t= 3sec

u = ?

We know by second Equaton of motion

S = u*t + 1/2*a*t^2

13.5 = u*3 + 1/2 *1 * 3^2          (9)

13.5 = u*3 + 9/2

13.5 - 4.5 = u*3

9 = u*3

u = 3m/s

******* Please Mark Brainliest ******************

Answer 2

Sorry I don’t know but you can search on google