aortiH
2020-11-06
pivonie8
Skilled2020-11-07Added 91 answers
For part a, if the ball is to pass through the hoophorizontally, its vert velocity at the hoop must be zero. The hoopis 4.9 m above the ground and the vert acc is -9.8m/s 2due to gravity. So... you can write:
Part b... this is like asking how many seconds will it take toreach the height of 4.9 m. Notice in part a we found the initialvertical speed is 9.8 m/s. Also, we know: vf= vi + at so 0 = 9.8 - 9.8t so t = 1 second
Part c... this gets a little tricky, but we have the info weneed. Consider that the speed he throws the ball at is 10.8m/s. This is a combination of the vert and horizcomponents. But we just figured out that the vert component must be9.8 m/s (part a) So if we make a right triangle of thecomponents and total, with x and y components as the sides and thetotal speed as the hypoteneuse, then we can use the pythagoreantheorem to find the x component of the speed he throws the ball at!Whew... in other words:
Now... he throws the ball with a horiz speed of 4.54 m/s ANDthe car is moving at 9.10 m/s. This means the ball is movingat 4.54 + 9.10 = 13.64 m/s toward the hoop,horizontally. We already said it would get to the hoop in onesecond, so... he must release the ball 13.64 m in front of thehoop.
Part d... you have to draw two triangles here. The vert leg inboth cases is 9.8 (the vert speed). The other leg and thehypoteneuse are different in both cases.
First: relative to car, horiz speed is 4.54m/s. So draw a triangle with legs 4.54 and 9.8. The hypoteneuse ofthis one will be the 10.8 that we used above. Now find the angle,using trig functions, between the horiz leg (4.54) and thehypoteneuse.
Second, relative to ground: horiz speed is 13.64, sodraw a triangle with legs 13.64 and 9.8. The hypoteneuse will besomething else that we havent calculated. Again, find the anglebetween the 13.64 side and the hypoteneuse.
Hope this helps, please rate me! Thanks!
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