Maximum Range Of A Projectile
Strings (SiPjAjk) = S7P4A41 Base Sequence = 12735 String Sequence = 12735 - 4 - 41
A projectile is fired with an initial velocity of 2000 ft/sec. How long does it take to reach a target at maximum range?
S7P4A41 (Motion - Linear).
Pj Problem of Interest (PPI) is of type motion. Problems of distance traveled are motion problems. The linearity is because the focus is on the range (horizontal distance) of projectile.
Maximum range = V2/32 (V is initial velocity). Also angle of fire = 450.
So, maximum range = (2000)2/32 = Vtcos45
So, t = √2(106)/(8 x 2000) = √2(62.5).
So, it take the projectile √2(62.5) secs to reach target at maximum range.
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