For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
The sled slides d=0.155 meters before rest.
Explanation:
m= 60 kg
V= 2 m/s
μ= 0.3
g= 9.8 m/s²
W= m * g
W= 588 N
Fr= μ* W
Fr= 176.4 N
∑F = m * a
a= (W+Fr)/m
a= 12.74m/s²
t= V/a
t= 0.156 s
d= V*t - a*t²/2
d= 0.155 m
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