# a ball is projected upwards from the top of the tower with the velocity 50 m/s making an angle 30 degree with the horizontal.The height of tower is 70m.After how many seconds from the instant of throwing will the ball reach the ground?

Sumit Majumdar IIT Delhi
10 years ago
Dear student,
The Total time after which the ball would reach the ground would be given by:
Regards
Sumit
Ananya Sharma
36 Points
9 years ago
But the correct answer is 7s. Pls. can u help me futher with my answer
Akshay
185 Points
9 years ago
@ananya: yes the answer is 7.
Let throwing point be (0,0). Horizontal be x-direction and vertical be y-direction
Initial velocity = (50cos30) i + (50sin30) j,
a = -10 j
s = ut + 0.5*a*t2,
s = (50*cos30*t) i + (50*sin30*t – 5*t2) j,
it will reach the ground when y-coordinate of ball is -70,
So, (50sin30*t – 5t2) = -70
or, 25*t-5*t2=-70, t2-5t-14=0
(t-7)(t+2)=0 => t=7
JohnCenaToTheRescue
20 Points
7 years ago
Why cant we calculate it by sumits formula.
Mukul
11 Points
6 years ago
Taking vertical downward motion of projectile from point of projection to ground ,we have
u=-50sin30°=-25m/s
a=+10m/s^2. ;s=70m ,t=?
Using  S=ut+1/2at^2
So, 70=-25×t+1/2×10×t^2
Or 5t^2-25t-70=0
Or t^2-5t-14=0
On solving,t=7s.{Here t^2 means t raise to power 2}
Rishi Sharma
4 years ago
Dear Student,

Let throwing point be (0,0). Horizontal be x-direction and vertical be y-direction
Initial velocity = (50cos30) i + (50sin30) j,
a = -10 j
s = ut + 0.5*a*t2,
s = (50*cos30*t) i + (50*sin30*t – 5*t2) j,
it will reach the ground when y-coordinate of ball is -70,
So, (50sin30*t – 5t2) = -70
or, 25*t-5*t2=-70
t^2-5t-14=0
(t-7)(t+2)=0
t=7

Thanks and Regards