Question

9. Two identical balls are simultaneously thrown towards each
other from points P and Q horizontally separated by 8 m,
and situated at heights 4 m and 8 m above the ground. One
ball is thrown from P horizontally with a speed of
8 ms, while the other is thrown downward with an initial
speed of v at an angle of 45° to the horizontal (figure). The
two balls collide in space, calculate (a) the initial speed of
ball thrown from the point Q.(b) coordinates of the point of
collision. g = 10 m s-2.​

Answer 1

Answer:

Both balls will travel vertically at a speed of 9.8 m/s very second. The firsball will travel at whatever speed it is throw at a constant velocity until gravity pulls it to the ground where it stops its motion. If the second ball drops vertically for one second, it will travel a distance of 9.8 meters. If the first ball is throw at a horizontal speed of 30 km/h from the same height as the second ball, it will travel a total distance of 30 km plus 9.8 km.

This is where you need to apply the Hypotenuse of a right triangle axiom . It is equal to the square root of the sum of the square of the sides. 30 km squared plus 9,.8 km squared equals 900 + 96.04 or 996.04. The square root of 996.04 is 31,56 km.

My suggested edit would have you not adding the horizontal and vertical numbers, but instead using those numbers as the legs of a right triangle and calculating the hypotenuse, which is the total velocity.t

Answer 2

Explanation:

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