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??

1 Answer
Feb 28, 2016

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 = 10ms^-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