Answer:
The second ball lands 1.5 s after the first ball.
Explanation:
Given;
initial velocity of the ball, u = 12 m/s
height of fall, h = 35 m
initial velocity of the second, v = 12 m/s
Time taken for the first ball to land;
determine the maximum height reached by the second ball;
v² = u² -2gh
at maximum height, the final velocity, v = 0
0 = 12² - (2 x 9.8)h
19.6h = 144
h = 144 / 19.6
h = 7.35 m
time to reach this height;
Total height above the ground to be traveled by the second ball is given as;
= 7.35 m + 35m
= 42.35 m
Time taken for the second ball to fall from this height;
total time spent in air by the second ball;
T = t₁ + t₂
T = 1.23 s + 2.94 s
T = 4.17 s
Time taken for the second ball to land after the first ball is given by;
t = 4.17 s - 2.67 s
T = 1.5 s
Therefore, the second ball lands 1.5 s after the first ball.
Complete Question
A 10 kg medicine ball is thrown at a velocity of 15 km/hr ( m/s) to a 50 kg skater who is at rest on the ice. The skater catches the ball and subsequently slides with the ball across the ice.
Calculate the kinetic energy after collision(in joules).
Answer:
Explanation:
From the question we are told that:
Mass of ball
Speed
Mass of Skater
Generally the equation for conservation of momentum is mathematically given by
Generally the equation for Kinetic energy is mathematically given by
Therefore kinetic energy K.E after collision is given as
Answer:
14.9 m /s
Explanation:
n = n° x v /[ v -v (s) ]
n is apparent frequency , n° is source frequency,v is speed of sound and v(s)
speed of source.
2300 = 2200 x 343 / [343 -v(s)]
v (s ) = 14.9 m /s.
Action force is a force acting in only one direction