Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction,
The distance between the buildings,
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
t = 0.707 s
Now,
When the policeman was chasing across:
The distance they will meet at:
9.57 - 2.0 = 7.57 m
Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
Answer:
0.00001 J
Explanation: