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
<span>two objects in contact with each other are the same temperature</span>
Answer:
Explanation:
mass of the ball = 146 g = 146 / 1000 = 0.146 kg
initial speed of the ball = 40.6 m/s
final speed of the ball = - 45.1 m/s
time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s
impulse, Ft = change in momentum = mv - mu = m (v-u)
F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N
Answer
Thanks
Explanation
B. A copper wire with rubber insulation
