Asked by Vishal
A basketball is launched with an initial speed of 12.0 m/s at 45 degrees from the ground.
The height of the basket y = 2.50m. From what distance x the ball is shot?
Note : The trajectory is like a basketball shot, the highest point the ball reaches is higher than the basket height.
The height of the basket y = 2.50m. From what distance x the ball is shot?
Note : The trajectory is like a basketball shot, the highest point the ball reaches is higher than the basket height.
Answers
Answered by
bobpursley
vertical:
2.5=vo*t*sintheta-4.9t^2
2.5=12*t*.707 - 4.9t^2
Horizontal:
d=12*.707 t
Ok, solve for t in the second equation
t=d/(12*.707)
put it in the first equation
2.5=12(d/12*.707)*.707- 4.9 (2d^2/144)
2.5=d-9.8 d^2/144
solve that quadratic for d. which is x in your problem description
2.5=vo*t*sintheta-4.9t^2
2.5=12*t*.707 - 4.9t^2
Horizontal:
d=12*.707 t
Ok, solve for t in the second equation
t=d/(12*.707)
put it in the first equation
2.5=12(d/12*.707)*.707- 4.9 (2d^2/144)
2.5=d-9.8 d^2/144
solve that quadratic for d. which is x in your problem description
Answered by
Vishal
The value of d is coming out to be 3.194 and 11.499. Can both be the answers? If not, what value to eliminate and why?
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