two identical balls are simultaneously thrown towards each other from point P and Q horizontally separated by 8 m and situated at height 4 and 8 metre above the ground one ball is thrown from P horizontally with a speed of 8 metre per second while other is thrown downward with an initial speed of V at an angle of 45 degree to the horizontal the two worlds collide in space calculate the initial speed of ball thrown from the point Q and the coordinates of point of collision.
Answers
Answer: Both balls will travel vertically at a speed of 9.8 m/s very second. The first ball 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.