Mechanics> a ball is projected upwards from the top ...

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?

maady manis , 10 Years ago

Grade 11

6 Answers

Sumit Majumdar

Last Activity: 10 Years ago

Dear student,
The Total time after which the ball would reach the ground would be given by:
$T=\frac{2usin\theta}{g}+\sqrt{\frac{2h}{g}}=\frac{2\times 50\times sin30^{\circ}}{10}+\sqrt{\frac{2\times 70}{10}}=5+3.74=8.74 sec$ Regards
Sumit

Ananya Sharma

Last Activity: 9 Years ago

But the correct answer is 7s. Pls. can u help me futher with my answer

Akshay

Last Activity: 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

Last Activity: 7 Years ago

Why can`t we calculate it by sumit`s formula.

Mukul

Last Activity: 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

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

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