Question

an athlete throws to a maximum distance of 80 m how long is it in air and to what height does it rise ? neglect the height of the athlete​

Answer 1

Answer:

The range of a projectile is R=u2sin2θg, and

the height of a projectile is H=u2sin2θ2g, where,

u is the magnitude of the initial velocity,

θ is the angle of projection, and,

g is the acceleration due to gravity.

⇒Rmax=u2g, at θ=45o.

The maximum range has been given as 80 m.

⇒ During this throw the angle of projection is 45o.

⇒u2g=80.

⇒ When the ball is thrown to the maximum distance possible, the maximum height is,

H=u2sin245o2g=(u22g)sin245o=802×12=20 m.

Edit:

My sincere thanks to P Muthu Samy who has pointed out that I had not included another 2 in the denominator in the last line. The error is now corrected.