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:
33.33 m/sec
Explanation:
A baseball travels 200 metes in 6 seconds,
what is the baseball’s velocity?
use the formula: velocity = distance over time
where (d) distance = 200 m
and (t) time = 6 sec.
plugin values into the formula:
v = d / t
= 200 m / 6 sec
= 33.33 m/sec.
therefore, the baseball's velocity is 33.33 m/sec
Momentum of the wagon increases by (200 x 3)
= 600 newton-sec
= 600 kg-m/sec
