Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s
Answer:
the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Explanation:
Since airplane is moving horizontally with constant speed v
so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane
so we can say that two object when dropped after some interval of time then they always lie in same vertical line
now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other
So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
To solve this question, we need to use the component method and split our displacements into their x and y vectors. We will assign north and east as the positive directions.
The first movement of 25m west is already split. x = -25m, y = 0m.
The second movement of 45m [E60N] needs to be split using trig.
x = 45cos60 = 22.5m
y = 45sin60 = 39.0m
Then, we add the two x and two y displacements to get the total displacement in each direction.
x = -25m + 22.5m = -2.5m
y = 0m + 39.0m
We can use Pythagorean theorem to find the total displacement.
d² = x² + y²
d = √(-2.5² + 39²)
d = 39.08m
And then we can use tan to find the angle.
inversetan(y/x) = angle
inversetan(39/2.5) = 86.3
Therefore, the total displacement is 39.08m [W86.3N]
