A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
R_"max" = 100 m
`Since `R_"max" = v^2/g => 100 = v^2/g`
Using equation of motion
`v^2 - u^2 = 2as`
Here, v = 0, a = -g, `s = R_"max"` = 100 m
`:.(0)^2 -u^2 = 2(-g) xx s`
`=> s = 1/2 u^2/g`
Since u = v
`:.s = 1/2 v^2/g = 1/2 xx 100 = 50 m`
Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
`R = (u^2sin2theta)/g`
`100 = u^2/g sin 90^@`
`u^2/g = 100` ..(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion
`v^2 - u^2 = -2gH`
`H=1/2xxu^2/g = 1/2xx100 = 50 m`