The rate of deceleration is equal to 25 m/s / 6 s = 25/6 m/s^2.
(a) The string should be angled in such a way that the horizontal acceleration is equal to 25.6 and the vertical acceleration is equal to gravitational acceleration. Using trigonometry,Angle from the vertical = arctan(25/6 / 9.8) = 23.03 degrees.
(b) Since the car is decelerating, one end of the string attached to the car is pulled back and because of the inertia of the object it is goes on the opposite direction and so the object moves toward the windshield.
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H=vt-16t²
vt-16t²=H
vt=16t²+H
v=(16t²+H)/t
v=16(t²/t)+H/t
v=16t+H/t
Answer: the answer would be: v=16t+H/t
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