Question

A particle is projected up with an initial velocity of
80 ft /sec . The ball will be at a height of 96 ft from the
ground 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 1

Answer:

After 1 second

Explanation:

S=ut+1/2at^2

96=80t+1/2×-10t^2

96=80t-5t^2

-96+80t-5t^2

T=2,3

So it is between 2 and 3

Answer 2

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 ).