Question

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

Answer 1

First of all u can't use this formula since its not including height...

Now, try n understand the question and prepare diagram..

we have, theta=30 degree , inital velocity=u=50m/s

so, horizontal component of u will be ux=u cos 30 and vertical component  is uy=usin30

now considering height.. since the ball is thrown from a tower of 70m and will reach the ground so the dispalcement will be -70m

now using s=ut+1/2 at2 {2 means sq} ie -70=50 sin30 t - 1/2* 10* t2 ((since a=g and we r considering vertical component ie height so                                                                                                         usin30))

solve it :  -70=25t - 5t2  => 5t2 - 25t -70 =0  => 5t + 10 =0 or t - 7 =0   => t= -2 or t= 7

but time cannot be in -ve so t = 7

ANS:-- t = 7 sec

is it correct?

Answer 2

Given :-

Velocity of ball = u = 50 Sin 30° = 25 ms-¹

Angle of Projection = Ø = 30°

Vertical distance = Sy = 70 m.

Taking upward motion as negative.

-Sy = ut - 1/2gt²

-70 = 25t - 1/2 × 10 t²

-70 = 25t - 5t²

5t² - 25t - 70 = 0

t² - 5t - 14 = 0

t² - 7t + 2t - 14 = 0

t(t-7) + 2(t-7) = 0

(t+2)(t-7) = 0

As we know that if product of two terms is zero then either first terms is equal to zero or second term is equal to zero.

t + 2 = 0

t = -2 s ( Not Valid)

or,

t - 7 = 0

t = 7 s

Hence,

Time Taken = t = 7 s.