Answer: B
Explanation:
Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.
The parameters given from the question are:
Mass M = 2kg
Height h = 20m
Let g = 9.8m/s^2
At the bottom of the incline plane, the object will experience maximum kinetic energy.
From conservative of energy, maximum K.K.E = maximum P.E
Maximum P.E = mgh
Maximum P.E = 2 × 9.8 × 20 = 392 J
But
K.E = 1/2mv^2
Substitute the values of energy and mass into the formula
392 = 1/2 × 2 × V^2
V^2 = 392
V = sqrt( 392 )
V = 19.8 m/s
V = 20 m/s approximately
Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
<span> Using conservation of energy
Potential Energy (Before) = Kinetic Energy (After)
mgh = 0.5mv^2
divide both sides by m
gh = 0.5v^2
h = (0.5V^2)/g
h = (0.5*2.2^2)/9.81
h = 0.25m
</span>