An object is dropped from a bridge.A second object is thrown downwards 2.00s later.They both reach the water 35.0m below at the same instant.What is the initial speed of the second object?
I will be very grateful if someone could help me out
Heres what I have so far
d=1/2(at^2)
d=35m
35=4.9t^2
t^2=(35/4.9)=7.14
t=2.67 sec
now the second object is thrown 2 sec later.It meets the first object at 2.67-2.00=0.67 sec
The distance travelled by the second object =1/2gt^2=1/2(9.8)(0.67^2)=2.2meters.
the distance travelled by the second object is 35-2.2=32.8 m
so initial velocity=(32.8/0.67)=49m/s
First Stone:
d = 0.5g*t^2 = 35 m.
4.9t^2 = 35
t^2 = 7.14
t = 2.67 s
Second Stone:
t = 2.67-2.0 = 0.67 s. to travel 35 m.
d = Vo*t + 0.5g*t^2
d = Vo*0.67 + 4.9*0.67^2 = 35
Vo*0.67 = 35 - 2.2 = 32.8
Vo = 49 m
Correction: Vo = 49 m/s.
Well, it seems like you've already done the calculations correctly! But let me add a little humor to your answer.
The initial speed of the second object is a whopping 49 m/s! That's faster than a cheetah running on roller skates! I hope it doesn't get a speeding ticket on its way down. Just imagine the second object yelling, "Wheeee!" as it zooms towards the water.
I hope this helps, and remember to always throw your objects with style and flair!
To find the initial speed of the second object, we can use the equation for distance traveled by an object in free fall:
d = 1/2 * g * t^2
Here, d represents the distance traveled (which is 35.0m), g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time it takes for the object to fall.
From your calculations, we know that the time it takes for the object to fall is 2.67 seconds. Therefore, we can substitute these values into the equation:
35.0 = 1/2 * 9.8 * (2.67)^2
Now, let's solve for the unknown term (1/2 * 9.8 * (2.67)^2):
1/2 * 9.8 * (2.67)^2 = 1/2 * 9.8 * (7.1289)
= 49.47903 m
So, the distance traveled by the second object is approximately 49.5 meters.
Now, let's calculate the distance traveled by the second object after it is thrown downwards 2.00 seconds later. We can use the same formula:
d = 1/2 * g * t^2
In this case, the time taken is 0.67 seconds (2.67 - 2.00 = 0.67 seconds). Substituting these values into the equation:
d = 1/2 * 9.8 * (0.67)^2
Simplifying this equation:
d = 1/2 * 9.8 * (0.4489)
d = 2.20019 m
So, the distance traveled by the second object after being thrown downwards 2.00 seconds later is approximately 2.2 meters.
To find the final answer, let's subtract this distance from the total distance (35.0m) in order to find the remaining distance traveled by the second object:
Remaining distance = Total distance - Distance traveled after being thrown
Remaining distance = 35.0 - 2.2
Remaining distance = 32.8 m
Now, we can calculate the initial velocity of the second object by dividing the remaining distance by the time it took for it to fall after being thrown:
Initial velocity = Remaining distance / Time
Initial velocity = 32.8 / 0.67
Initial velocity ≈ 48.89 m/s
Therefore, the initial speed of the second object is approximately 48.89 m/s.