Question

If a projectile is shot at an angle of `pi/4` and at a velocity of `15 m/s`, when will it reach its maximum height??

Answer 1

2.12s

Explanation:

Velocity of projection`,v=15ms^-1`
Angle of projection`,alpha=pi/4`
Vertical component of vel. of projection`,vsinalpha=15*sin(pi/4)=15/sqrt2ms^-1`
Applying equn for vertical motion under gravity,
at maximum height (H) the final vertical vel. being ZERO, we can write
`0^2=(vsinalpha)^2-2*g*H`
`=>H=(vsinalpha)^2/(2g)=15^2/(2^2*10) =5.625m`, [taking g = 10`ms^-2`]
If time of reaching Maximum height be T then
`0=vsinalpha*T-1/2gT^2`
`=>T=(2vsinalpha)/g=2*15/(10sqrt2)=2.12s`

Calculation of height not wanted here