Question

A boy stands on a fishing pier and tosses a pebble into a pond. The pebble is thrown from a height of 2.50 m above the water surface with a velocity of 4.80 m/s at an angle of 60.0° above the horizontal. As the pebble strikes the water surface, it immediately slows down to exactly half the speed it had when it struck the water and maintains that speed while in the water. After the pebble enters the water, it moves in a straight line in the direction of the velocity it had when it struck the water. If the pond is 2.90 m deep, how much time elapses (in s) between when the pebble is thrown and when it strikes the bottom of the pond?

s

Answer 1

let
vo = 4.80 m/s
theta = 60 degrees
h = 2.50 m
vox = 4.8*cos(60) = 2.4 m/s
voy = 4.8*sin(60) = 4.157 m/s

let t1 is the time taken for the pebble to hit the water.

use,

-h = vo*t1 + (1/2)*(-g)*t1^2

-2.5 = 4.157*t1 + (1/2)*(-9.8)*t1^2

==> t1 = 1.255 s

when the pebble hits,
vx = vox = 2.4 m/s
vy = voy - g*t1

= 4.157 - 9.8*1.255

= -8.14 m/s

just before hitting the water the speed of the pebble,
v = sqrt(vx^2 + vy^2)

= sqrt(2.4^2 + 8.14^2)

= 8.50 m/s

aafter hitting the water the speed of the pebble, v' = v/2

= 8.50/2

= 4.25 m/s

direction : theta = tan^-1(vy/vx)

= tan^-1(8.14/2.4)

= 73.6 degrees below horizontal.

distance travelled in water, d = 2.9/sin(73.6)

= 3.02 m

so, time taken for the pebble to reach bottom of the pool, t2 = d/v'

= 3.02/4.25

= 0.7106 s

total time taken, t = t1 + t2

= 1.255 + 0.7106

= 1.96 s <<<<<<<<<<<--------------------Answer