After time 't' the height ‘y ’ of projectile is y = 8t - 5t^(2) and horizontal distance x = 6t if g=10 ms^(-2)1 , velocity of projectile at this instant is :

After time 't' the height ‘y ’ of projectile is y = 8t - 5t^(2) an

1.	After time 't' the height ‘y ’ of projectile is y = 8t - 5t^(2) and horizontal distance x = 6t if g=10 ms^(-2)1 , velocity of projectile at this instant is :
Answer» <html><body><p>10m/s<br/>8m/s<br/>6m/s<br/>4m/s</p>Solution :During <a href="https://interviewquestions.tuteehub.com/tag/projectile-1169321" style="font-weight:bold;" target="_blank" title="Click to know more about PROJECTILE">PROJECTILE</a>’s <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a><br/>`x=ucosthetat`<br/>`6t=ucosthetatimplies ucostheta=6impliesu_(x)=6ms^(-1)`<br/>Similarly `y=usinthetat-1/2g t^(2)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/8t-340101" style="font-weight:bold;" target="_blank" title="Click to know more about 8T">8T</a> - 5t^(2)=usinthetat-1/2g t^(2)`<br/>`implies usintheta=8impliesu_(y)=<a href="https://interviewquestions.tuteehub.com/tag/8ms-1930770" style="font-weight:bold;" target="_blank" title="Click to know more about 8MS">8MS</a>^(-1)`<br/>Now `u=sqrt(u_(x)^(2)+u_(y)^(2))=10ms^(-1)`</body></html>

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After time 't' the height ‘y ’ of projectile is y = 8t - 5t^(2) and horizontal distance x = 6t if g=10 ms^(-2)1 , velocity of projectile at this instant is :

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