Question #64bde

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Answer 1 · 2017-08-22T17:20:41

Option #(2)# is correct.

Range #R# of a projectile is given by

#R=(v_0^2sin2theta)/g#
For above, for a given angle and constant value of #g# we conclude that:

#=>Rpropv_0^2#

Now, if velocity #v^2# is double of #v_0#, then range #R'# becomes #4# times of range #R#, i.e;

#R'=4R#

Four times means #400%#,

#=>R'=400%R#

Now, percentage change in range can be calculated as

Change#=R'-R=400%R-100%R=300%R#

#implies# percentage change is #300%#.

Hecne, option #(2)# is correct.