A ball whose Kinetic Energy is E is projected at an angle of `45^(@)` to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

Answer

A, B, C

Explanation

Given, `g= 9.8 m/s^2 = 32.2 `ft/s^2``
` 40` yd= 120 t
Horizontal range
x= 120 ft
u= 64 ft/s
theta= 45^0
We know that horizonta
range
`x=u cos theta t
rarr t= x/(u cos theta)
= 120/(64 cos 45^0)= 2.65 sec
y= sin theta (t)- 1/2 gt^2
= 64(1/sqrt(2))(2.65)- 1/2 (32.2)(2.65)^2
= 7.08 ft which is less than the 8th` of goal post In tiem 2.65 s the ball travels horizontl distnce 120 ft ( 40 yd) and vertical height 7.08 ft which is less than 10 ft. The ball will reach the goal post.