Abatter hits the baseball a with an initial velocity of v0 = 110 ft/sec directly toward fielder b at an angle of 23° to the horizontal; the initial position of the ball is 2.2 ft above ground level. fielder b requires 0.44 sec to judge where the ball should be caught and begins moving to that position with constant speed. because of great experience, fielder b chooses his running speed so that he arrives at the “catch position” simultaneously with the baseball. the catch position is the field location at which the ball altitude is 8.4 ft. determine the velocity of the ball relative to the fielder at the instant the catch is made.
(a) The net force is: 271.7 (i) N
(b) The acceleration is: 3.8 m/s²
(c) The net force increases
In order to answer the questions you have to apply Newton's Second Law:
Where F is the force, m is the mass and A is the acceleration.
Applying it for the x-axis and y-axis (Where the x-axis is in the slope's direction and the y-axis is perpendicular to it)
x-axis: Wx =mA
y-axis: N-Wy=0 (because the velocity in the y-axis is constant)
Wx and Wy are the components of the weight vector.
You have to obtain the components of the weight. The slope is at an angle of 23° with the horizontal and the weight is perpendicular to the horizontal, therefore the angle formed by the weight and the slope is: 90-23=67°
Applying trigonometric identities for a right triangle to obtain the components of the weight:
Wx= mgCos(67°)=(71)(9.8)Cos(67°)=271.9 N
Wy= mgSin(67°)=(71)(9.8)Sin(67°)=640.5 N
Therefore the net force is the sum of the net force in x and the net force in y (which is zero)
F= 271.7 N
Where i is the unit vector.
The acceleration is:
A=Wx/m=271.7/71 = 3.8 m/s²
For part (c):
If the ski slope becomes steeper it means that the angle increases. The net force is: mgCos(90-θ) which is equal to mgSin(θ). Therefore, if the angle increases then Sin(θ) increases and the force increases.
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