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coldgirl [10]
3 years ago
14

An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t

he same instant. What was the initial speed of the second object
Physics
1 answer:
pashok25 [27]3 years ago
8 0

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

x = v_0 t +\frac{1}{2}at^2

48.1 = 0*t + \frac{1}{2} (9.8)t^2

t = 3.13s

The second object is thrown downward at one second later and it meets the first object at the water is

t' = 3.13 -1.48

t' = 1.65s

The distance of the object will travel due to free fall acceleration is

x = v_0 t+\frac{1}{2} at^2

x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2

x = 13.34m

The distance of the object will travel due to its initial velocity is

v_0 = \frac{d_0}{t}

d_0 = v_0 t

48.1-13.34 = v_0 (1.65)

v_0 = 21.06m/s

Therefore the initial speed of the second object is 21.06m/s

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