An object is dropped from a height of 49 meters. The object takes 3.2s to hit the ground. The final velocity of the object when it hits the ground is -31 m/s
d = -(1/2) g t^2
-49 = -(1/2)(9.8)t^2
t^2 = 10
t = 3.16 seconds
v = -gt = -9.8*3.16 = -31 m/s
Well, I must say that object really knows how to make a grand entrance! It's like it's auditioning for a stunt double in an action movie. Dropping from a height of 49 meters and reaching a final velocity of -31 m/s in just 3.2 seconds? That's pretty impressive! It sounds like this object is in a bit of a hurry to get to the ground. Just hope it doesn't miss its mark and land on someone's foot!
To find the initial velocity of the object, we can use the equation of motion for a falling object. The equation is given by:
s = ut + (1/2)gt^2
Where:
- s is the distance traveled (in this case, the height) (49 meters)
- u is the initial velocity of the object (which we need to find)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken (3.2 seconds)
Rearranging the equation to solve for u, we get:
u = (s - (1/2)gt^2)/t
Substituting the given values, we have:
u = (49 - (1/2)(-9.8)(3.2^2))/3.2
Calculating the denominator first, we have:
(1/2)(-9.8)(3.2^2) = -1/2 * 9.8 * 10.24 = -49 * 10.24 = -502.76
Substituting this in the numerator, we have:
u = (49 - (-502.76))/3.2
= (49+502.76)/3.2
= 551.76/3.2
= 172.425
Therefore, the initial velocity of the object is approximately 172.425 m/s.