The answer is:
The velocity with which the ball can be thrown to have a maximum range of 20 meters is equal to 14 m/s.
[tex]u=14\frac{m}{s}[/tex]
To solve the problem and find the velocity, we need to isolate it from the equation used to calculate the maximum horizontal range.
We have the equation:
[tex]R=\frac{u^{2} }{g}[/tex]
Where,
R is the maximum horizontal range.
u is the initial velocity.
g is the gravity acceleration.
Also, from the statement we know that:
[tex]R=20m\\g=9.8\frac{m}{s^{2} }[/tex]
So, using the given information, and isolating, we have:
[tex]R=\frac{u^{2} }{g}[/tex]
[tex]R*g=u^{2}[/tex]
[tex]u^{2}=R*g=20m*9.8\frac{m}{s^{2} }=196\frac{m^{2} }{s^{2} }\\\\u=\sqrt{196\frac{m^{2} }{s^{2}}}=14\frac{m}{s}[/tex]
Hence, we have that the velocity with which the ball can be thrown to have a maximum range of 20 meters is equal to 14 m/s.
[tex]u=14\frac{m}{s}[/tex]
Have a nice day!
Answer:
The velocity with which a ball must be thrown to have a maximum range of 20 m is 14 m/s.
Note that this problem means to find the magnitude of the velocity and not the direction (it is implicit in the formula that the angle of the launch is 45°).
Explanation:
You just must use the given equation for the maximum horizontal range of a projectile and solve for u which is the unknwon:
Solve for u:
Take square root from both sides:
