UPSEE Solved Model Paper 7

Given:
Initial velocity of the ball: u = 15 m/s
Initial angle with the horizontal θ = 30°
We have to find the time of flight of the ball
Corresponding figure is shown below:

The time taken by the ball to reach the point 'C' from O is
T =

2usinθ

g

⇒ T =

2×15m∕s×sin30°

10m∕s2

= 1.5 s
Neglecting friction. the ball attains the same velocity u, at the point 'C' (which is at the same
level with point O) (Using law of conservation of mechanical energy).
Considering Y-motion of the ball from point C' onwards, we have
h - 10.5 m + 2 m = 12.5 m
a = g = 1 0 m∕s2
Using second equation of motion: S = ut +

1

2

at we obtain
h = (usinθ)t+

1

2

gt2
⇒ 12.5 m = (15 m/s × sin 30° × t) + (

1

2

×10×t2)
⇒ 5t2+7.5t−12.5 = 0
⇒ (5t + 12.5) (t - 1) = 0
⇒ t = 1 s , t = −

12.5

5

s
[Initial velocity in Y - direction is uy = u sin θ]

Out of the two values , t = 1 s is the possible physical solution of the above equation
Hence, the total time taken by the ball to reach the ground is
t' = T + t = 1.5 s + 1 s = 2.5 s
Hence, option 'C' is correct