Question

A ball thrown vertically up from the top of a tower
reaches the ground in 12 s. Another ball thrown ver-
tically downwards from the same position with the
same speed takes 4 s to reach the ground. Find the
height of the tower. (Take g = 10 ms^-2)​

Answer 1

Answer:

Ball is thrown from the top of tower, ⇒height=−h,

For the first case, velocity = u , Time t=12s,

Using s=ut+

2

1

at

2

⇒−h=u×12−

2

1

×10×12

2

⇒−h=12u−720 .....1

For the second case, velocity = −u , Time t=4s,

Using s=ut+

2

1

at

2

⇒−h=−u×4−

2

1

×10×4

2

⇒−h=−4u−80 .....2

Equating equations 1 and 2,

⇒12u−720=−4u−80 ⇒u=40ms

−1

Put the value of u=40ms

−1

in equation 1,

⇒−h=12×40−720 ⇒h=240m

Answer 2

Ball is thrown from the top of tower, ⇒height=−h,

For the first case, velocity = u , Time t=12s,

Using s=ut+

2

1

at

2

⇒−h=u×12−

2

1

×10×12

2

⇒−h=12u−720 .....1

For the second case, velocity = −u , Time t=4s,

Using s=ut+

2

1

at

2

⇒−h=−u×4−

2

1

×10×4

2

⇒−h=−4u−80 .....2

Equating equations 1 and 2,

⇒12u−720=−4u−80 ⇒u=40ms

−1

Put the value of u=40ms

−1

in equation 1,

⇒−h=12×40−720 ⇒h=240m

here you go