To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.
Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds
Horizontal distance
=0.335s*1.2m/s
=0.402 meters
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Answer:
20 m/s
30 m/s
Explanation:
Given:
v₀ = -10 m/s
a = -9.8 m/s²
When t = 1 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (1 s)
v = -19.8 m/s
When t = 2 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (2 s)
v = -29.6 m/s
Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.
Answer:
Explanation:
Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:
F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to
F = f which states that the applied Force equals the frictional force, choice a.
A bowler throws a bowling ball that rolls down the lane, im like 98% sure it’s that one