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:
Explanation:
given,
F = 14.1 i + 0 j + 5.1 k
displacement = 6 m
Assuming block is moving in x- direction
we know,
dW = F dx
![W = F[x]_0^6](https://tex.z-dn.net/?f=W%20%3D%20F%5Bx%5D_0%5E6)
hence, work done by the force is equal to
Explanation:
a. moving a 145 kg aluminum sculpture across a horizontal steel platform
b. pulling a 15 kg steel sword across a horizontal steel shield
c. pushing a 250 kg wood bed on a horizontal wood floor
d. sliding a 0.55 kg glass amulet on a horizontal glass display case
