You need to divide the motion into its component: vertical and horizontal motion.
The time taken to fall vertically from the cliff is equal to the time taken to move horizontally.
Using the vertical component, which is an accelerated motion with an initial velocity equal to zero, we can solve for t:
h = 1/2 · g · t²
t = √(2·h / g)
= √(2·50 / 9.8)
= 3.2 s
Horizontally, it is a constant motion:
d = v · t
= 20 · 3.2
= 64 m
The ball will strike the ground at a distance of 64 meters from the cliff.
Answer:
a) distance d = 293.36ft
b) acceleration a = 14.67ft/s^2
Explanation:
Acceleration is the change in velocity per unit time.
a = ∆v/t ....1
Given;
Initial velocity vi = 30mph × 5280ft/mile × 1/3600s/h
vi = 44ft/s
Final velocity vf = 70mph × 5280ft/mile × 1/3600s/h
vf = 102.67ft/s
time = 4.0s
From equation 1, acceleration is;
a = ∆v/t = (102.67-44)/4 = 14.67ft/s^2
Distance travelled can be given as;
d = ut + 0.5at^2 .....2
u = 44ft/s
t = 4
a = 14.67ft/s^2
Substituting into the equation 2
d = 44(4) + 0.5(14.67×4^2)
d = 293.36ft
Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
Answer:
cause Ice is lighter than sunglasses
D = V1( t ) + 1/2g( t )^2
50m = 0m/s( t ) + 1/2(9.8m/s^2)*( t )^2
V1*t cancels out
50m = (4.9m/s^2)*(t)^2
50m/(4.9m/s^2) = t^2
Metres unit cancels out so we are left with s^2
10.204s^2 = t^2
Square root both sides to cancel out square
t = 3.19 s
