(a) Neglecting air resistance, how fast is the object traveling when it strikes the ground at the end of 2.0 s? First used standard British units and then express the speed in mi/h for a familiar comparison
v=gt
21.56
To calculate the speed at which the object strikes the ground, we can utilize the equations of motion. In this case, we need to determine the final velocity of the object when it reaches the ground after 2.0 seconds.
Since we are neglecting air resistance, the only force acting on the object is gravity. Thus, we can use the equation of motion for vertical motion:
v = u + gt
Where:
v = final velocity
u = initial velocity (which we assume to be zero since the object was dropped)
g = acceleration due to gravity (approximately 32.17405 ft/s²)
t = time taken
Plugging in the given values, we have:
v = 0 + (32.17405 ft/s²) × (2.0 s)
v = 64.35 ft/s
Now, let's convert the speed to miles per hour for a more familiar comparison. We know that 1 mile equals 5280 feet, and 1 hour is equal to 3600 seconds. Therefore:
Speed in mi/h = (Speed in ft/s) × (1 mi/5280 ft) × (3600 s/1 h)
Plugging in the values:
Speed in mi/h = (64.35 ft/s) × (1 mi/5280 ft) × (3600 s/1 h)
Speed in mi/h ≈ 43.91 mi/h
So, neglecting air resistance, the object would be traveling at approximately 43.91 mi/h when it strikes the ground after 2.0 seconds.