Question

31. From the top of a tower of height 40 m, a ball is projected
upwards with a speed of 20 ms at an angle of 30° to the
horizontal. If g = 10 ms?, after how long will the ball hit
the ground?



(1) 15
(2) 25
(3) 35
4) 4s​

Answer 1

Solution :-

please refer the attachment for better understanding ,

As per the given data ,

• Height of the tower(y) = 40 m
• Angle of projection(Ф) = 30°
• Acceleration due to gravity(g) = 10 m/s²

The ball moves in a projectile from point A to B hence , the time of flight is given by ,

➝ T = 2usinФ / g

➝ T = 2 x 20 x sin 30 / 10

➝ T = 2 x 20 x 1 / 20

➝ T = 2 s

Now while moving from point B to C

By applying second equation of motion ,

➝ y = uy + at² / 2

➝ y = - u cos (60 )t + 10 x t ² / 2

➝ 40 = - 20 x 1 / 2 x t+ 5 t²

➝ 40 = - 10t + 5 t²

➝ 5t ² - 10 t - 40 = 0

➝ t ² - 2t - 8 = 0

➝ t² - 4t + 2t - 8 = 0

➝ t ( t- 4 ) + 2 ( t- 4) = 0

➝ (t + 2)( t - 4 ) =0

➝ t = -2 s  or t = 4 s

We know that time cannot be negative

Hence ,

➝ t = 4 s

Hence the total time required by the ball to hit the ground

=T + t

= 2 + 4

= 6 s

The total time required by the ball to hit the ground is 6s