A soccer player kicks a rock horizontally off the a 40 m high cliff into the pool of water. if the player hears the sound of the splash 3 sec., what is the initial speed given to the rock? Assume the speed of the sound in air to be 343 m/s.
Time for rock to fall 40 meters?
d = (1/2) a t^2
40 = (1/2)(9.8) t^2
t^2 = 8.16
t = 2.86 seconds
so the sound traveled for 3-2.86 = .143 seconds
so the rock hit 343*.143 = 49 meters away
that is hypotenuse of angle
with leg 40 and other leg is horizontal distance d from bottom of cliff to splash
d = sqrt (49^2 -40^2)
d = 28.3 meters
so rock hit 28.3 meters from cliff base in its 2.86 meters in the air
speed = distance/time = 28.3/2.86 = 9.9 m/s
sample answer step by step
I am sureprisdd
To solve this problem, we can use the concept of projectile motion. Let's break the problem down into two parts:
1. Calculate the time taken for the rock to fall from the cliff to the water.
2. Use the time calculated in step 1 to determine the initial speed of the rock.
Step 1: Calculate the time taken for the rock to fall from the cliff to the water.
We can use the equation for the vertical motion of a projectile:
h = (1/2) * g * t^2
where h is the height (40 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken for the rock to fall.
Using this equation, we can rearrange it to solve for t:
t^2 = (2h) / g
t^2 = (2 * 40) / 9.8
t^2 = 64.4898
t ≈ 8.03 seconds (rounded to two decimal places)
Step 2: Use the time calculated in step 1 to determine the initial speed of the rock.
We'll assume that the initial horizontal velocity of the rock is the same as the final horizontal velocity (since no horizontal forces act on the rock).
Using the formula:
distance = speed * time
we can determine the distance traveled by the sound:
distance = speed of sound * time
343 m/s * 3 sec = 1029 m
We can then use this distance to calculate the initial velocity of the rock:
distance = initial velocity * time
1029 m = initial velocity * 8.03 sec
Solving for the initial velocity, we get:
initial velocity = 1029 m / 8.03 sec ≈ 128 m/s
Therefore, the initial speed given to the rock is approximately 128 m/s.