Homework Help: physics

Posted by applebottom on Friday, October 8, 2010 at 7:05pm.

A basketball is thrown at 45° to the horizontal. The hoop is located 4.5m away horizontally at a height of 1m above the point of release. What is the required initial speed?

physics - bobpursley, Friday, October 8, 2010 at 7:10pm
Horizontal
Vcos45*t=4.5
Vertical:
Vsin45*t-4.9t^2=1

solve for t in the first equation (in terms of V), put that into the second, and solve for V
physics - applebottom, Friday, October 8, 2010 at 7:15pm
how do u solve v?
physics - bobpursley, Friday, October 8, 2010 at 7:16pm
Put t=4.5/Vcos45 into the second equation, you only have V as the unknown. It is a quadratic, use the quadratic equation.
physics - Anonymous, Wednesday, October 17, 2012 at 7:14pm
Not good explanation
physics - Anonymous, Wednesday, October 17, 2012 at 7:14pm
Not good explanation

Answer this Question

Related Questions

Physics - A basketball player throws the ball at a 36° angle above the ...
Physics please help desprate - A basketball player throws the ball at a 38° ...
Physics please help!! - A basketball player throws the ball at a 38° angle ...
Physics - A winning basketball shot in the last second of a game is launched ...
Physics - A basketball is thrown horizontally with an initial speed of 4.20 m/s...
Physics - Your teacher tosses a basketball. The ball gets through the hoop (...
physics - 1. A basketball player passes the ball to his teammate at an angle of ...
physic - Craig launches the basketball at an angle of 65.0 degrees above the ...
Physics - A player is shooting the basketball 10 meters from the hoop. He ...
physics - A basketball player is standing on the floor 10.0 m from the basket as...

For Further Reading

First Name:
School Subject:
Answer: