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

Answer:

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