Question

A ball is thrown vertically upwards with a velocity of 15 m s-' from the top of the building of height
50 m.
(a) How long will it take to reach the base of the tower?
(b) With what velocity it hits the ground?​

Answer 1

a) We have to find out time taken by ball to reach at point B which is of the same level of base of base of tower. [ refer attachment for diagram ]

When ball is thrown upward then at point A it's velocity will be 0m/s because it is the highest point and at height point velocity is always zero.

So we have :-

• initial velocity at C = 15 m/s
• final velocity at point A = 0m/s
• acceleration = g = 10m/s²

Now by using formula v= u+at we will find time taken by ball to reach from point C to A.

$\implies \sf \: v = u + at$

where :-

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time

Let time taken by ball to reach from point C to A = t_1

$\implies \sf \: 0 = 15 + ( - 10)t_1$

$\implies \sf \: - 15 = - 10t_1$

$\implies \sf \: \dfrac{ - 15}{ - 10} = t_1$

$\implies \sf \: \dfrac{ 15}{ 10} = t_1$

$\implies \sf \: { 1.5} \: s = t_1$

Now we will find distance between point C and A by using v²-u²= 2as

Where :-

• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance

$\sf \implies {v}^{2} - {u}^{2} = 2as$

$\sf \implies {0}^{2} - {(15)}^{2} = 2( - 10)s$

$\sf \implies {(15)}^{2} = 20s$

$\sf \implies \dfrac{{15} \times 15}{20} = s$

$\sf \implies \dfrac{{3} \times 15}{4} = s$

$\sf \implies \dfrac{45}{4} = s$

$\sf \implies 11.25 \: m = s$

Now use s = ut+1/2 at² to find time taken to reach the ball from A to B.

At point A velocity is zero.

Total distance between A and B = 11.25+50 = 61.25 m

$\sf \implies \: s = ut + \dfrac{1}{2} a {t}^{2}$

$\sf \implies \: 61.25 = (0 \times t)+ \dfrac{1}{2} (10) {t}^{2}$

$\sf \implies \: 61.25 = 5 {t}^{2}$

$\sf \implies \: \dfrac{61.25}{5} = {t}^{2}$

$\sf \implies \: \dfrac{6125}{500} = {t}^{2}$

$\sf \implies \: \dfrac{1225}{100} = {t}^{2}$

$\sf \implies \: \sqrt{ \dfrac{1225}{100} } = {t}$

$\sf \implies \dfrac{35}{10} = t$

$\sf \implies {3.5} \: s = t$

Total time taken = t_1 + t

Total time taken = 1.5+3.5 = 5 seconds

(b) Now we have to find out velocity with which ball hit the ground.

=> So , at point A velocity = 0 m/s ( it is considered as initial velocity)

=> Time taken for ball to travel from point A to B = 3.5 sec

=> Distance between A to B = 61.25 m

To find final velocity we will use the formula v = u+at

$\sf \implies v = u + at$

$\sf \implies v = 0 +( 10 \times 3.5)$

$\sf \implies v = 35 \frac{m}{s}$

velocity with which ball hit the ground = 35 m/s