A projectile is thrown to have its max height as H. for what height the ratios of time of being there will be alpha??
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The trajectory for projectile motion is symmetric about the point of maximum height. The projectile covers the same horizontal distance reaching its maximum height as it does falling from its maximum height back to the ground. It takes the projectile the same amount of time reaching its maximum height as it does to fall from the maximum height back to the ground.
Time of flight is total time taken by projectile.
Using equation for vertical motion for projectile we have
H= ut −12gt2gt2−2ut +2H =0t =−b±b2−4ac√2at1=2u+4u2−8gH√2g=2u+2u2−2gH√2g=2u+2v2√2g=ug (v =0 at maximum height)t2=2u−4u2−8gH√2g=ugTherefore time of flightT =t1+t2= 2ugMaximum height H =u22g=(gT/2)22g=(g2T2)8g=(gT2)8=g(t1+t2)28
Regards
The trajectory for projectile motion is symmetric about the point of maximum height. The projectile covers the same horizontal distance reaching its maximum height as it does falling from its maximum height back to the ground. It takes the projectile the same amount of time reaching its maximum height as it does to fall from the maximum height back to the ground.
Time of flight is total time taken by projectile.
Using equation for vertical motion for projectile we have
H= ut −12gt2gt2−2ut +2H =0t =−b±b2−4ac√2at1=2u+4u2−8gH√2g=2u+2u2−2gH√2g=2u+2v2√2g=ug (v =0 at maximum height)t2=2u−4u2−8gH√2g=ugTherefore time of flightT =t1+t2= 2ugMaximum height H =u22g=(gT/2)22g=(g2T2)8g=(gT2)8=g(t1+t2)28
Regards
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