# a projectile is fired vertically at an angle theta from ground with velocity v. speed of projectile when it is moving at an angle delta from horizontal is ​

Dear student

when projectile fired vertically with an angle $\theta$ from the ground it's horizontal and vertical component of velocity are given as  . Also there is no acceleration in the horizontal so the horizontal component of velocity does not change during the projectile motion.
Let after a time t the projectile makes angle $\delta$ from the horizontal and let the velocity is u then the horizontal and vertical component of velocity at this time is given by
${u}_{x}={v}_{x}=u\mathrm{cos}\delta =v\mathrm{cos}\theta$
since the acceleration in the vertical direction is g so the vertical velocity is given by
${u}_{y}={v}_{y}-gt\phantom{\rule{0ex}{0ex}}{u}_{y}=v\mathrm{sin}\theta -gt$
But since u makes the angle $\delta$ from the horizontal so

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