Answer:
U = √Rg/sin2θ
Explanation:
Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.
Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.
According to the range formula,
R = U²sin2θ/g
Cross multiplying we have;
Rg = U²sin2θ
Dividing both sides by sin2θ, we have;
U² = Rg/sin2θ
Taking the square root of both sides we have;
√U² = √Rg/sin2θ
U = √Rg/sin2θ
Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ
Q3. (a) 0m/s, as they are asking for initial velocit.
(b)(i) The paper has a large surface area or weighs less than the coin,thus,falls smoothly.
(ii)The coin has more mass than the paper.
(c) They fall at the same acceleration and hit the bottom at the same time.
Their mass doesn't matter in the vacuum.
Answer:
100
Explanation:
Momentum (P) = Mass (M) × Velocity (V)
Momentum = 25kg × 4.0m/s
Momentum = 100kg m/s
Active Optics.
Hope that helps, Good luck! (:
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
