A particle is projected up with an initial velocity of80 ft /sec . The

1.	A particle is projected up with an initial velocity of80 ft /sec . The ball will be at a height of 96 ft from theground after(a) 2.0 and 3.0 sec (b) Only at 3.0 sec(c) Only at 2.0 sec (d) After 1 and 2 sec
Answer» Solution : Initial VELOCITY, u = 80 ft/sec Height, h = 96 ft ACCELERATION DUE to gravity, g = 32 ft/sec^2 Now, *Using 2nd Equation of motion, h = ut* + 1/2 gt^2 96 = 80 t - 1/2 ( 32 )t^2 [ - sign for above the ground ] 96 = 80T - 16 t^2 t^2 - 5t + 6 = 0 t^2 - 3t - 2T + 6 = 0 t ( t - 3 ) - 2( t - 3 ) = 0 ( t - 2 ) ( t - 3) = 0 t = 2, 3 Hence , *Ball will be at a height of 96 ft after 2.0 and 3.0 seconds.* Option ( A ).

A particle is projected up with an initial velocity of80 ft /sec . The ball will be at a height of 96 ft from theground after(a) 2.0 and 3.0 sec (b) Only at 3.0 sec(c) Only at 2.0 sec (d) After 1 and 2 sec

Answer»

Solution :

Initial VELOCITY, u = 80 ft/sec

Height, h = 96 ft

ACCELERATION DUE to gravity, g = 32 ft/sec^2

Now, Using 2nd Equation of motion,

h = ut + 1/2 gt^2

96 = 80 t - 1/2 ( 32 )t^2 [ - sign for above the ground ]

96 = 80T - 16 t^2

t^2 - 5t + 6 = 0

t^2 - 3t - 2T + 6 = 0

t ( t - 3 ) - 2( t - 3 ) = 0

( t - 2 ) ( t - 3) = 0

t = 2, 3

Hence , Ball will be at a height of 96 ft after 2.0 and 3.0 seconds.

Option ( A ).

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A particle is projected up with an initial velocity of80 ft /sec . The ball will be at a height of 96 ft from theground after(a) 2.0 and 3.0 sec (b) Only at 3.0 sec(c) Only at 2.0 sec (d) After 1 and 2 sec​