Question

5. A ball is gently dropped from height of 5 m. If its velocity increases uniformly at
the Rate of 10 ms, after what time will it strike the ground?
(a) 3s, (b) 2 s (c) 1s, (d) 0.5 s.​

Answer 1

Answer:

After 1 s, the ball will strike the ground

Explanation:

Given,

s = 5 m

a = 10 m/s^2

u = 0 m/s

Formula,

s = ut + 1/2 at^2

5 = 0 + 1/2 10 t^2

t^2 = 1

t = 1 s

Answer 2

Diagram:

$\setlength{\unitlength}{1mm}\begin{picture}(7,2)\thicklines\multiput(7,2)(1,0){55}{\line(3,4){2}}\multiput(35,7)(0,4){12}{\line(0,1){0.5}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{12}}\put(37,7){\large\sf{v = 10 mps}}\put(37,55){\large\sf{u = 0 m/s}}\put(21,61){\textsf{\textbf{Ball}}}\put(43,40){\line(0, - 4){28}}\put(43,34){\vector(4){18}} {\pmb{\sf{BrainlyButterfliee}}}\put(24, - 3){\large\sf{ \sf g = + 10 m/s^2 }}\put(48,30){\large\sf{H = 5 \: m}} \: \: \: {\large\sf{Time = ?}}\end{picture}$

Provided that:

• Height = 5 metres
• Final velocity = 10 mps
• Initial velocity = 0 mps
• g = +10 m/s sq.

Don't be confused!

• Initial velocity cames as zero because a ball is gently dropped from height means it drops from rest.

• g denotes acceleration due to gravity and we also know that it's universal value is 9.8 m/s sq. But here we are taking it as 10 m/s sq. as approx. And it is positive because the ball is going downwards.

To calculate:

• Time taken

Solution:

• Time taken = 1 second

Using concept(s):

→ We can solve this question by using either first equation of motion or second equation of motion.

Using formula(s):

→ 1st eqⁿ => v = u + at

→ 2nd eqⁿ => s = ut + ½ at²

Required solution:

By using second equation of motion:

→ s = ut + ½ at²

→ 5 = 0(t) + ½ × 10(t)²

→ 5 = 5t²

→ 1 = t²

→ √1 = t

→ 1 = t

→ t = 1 second

→ Time = 1 second

By using first equation of motion:

→ v = u + at

→ 10 = 0 + 10(t)

→ 10 = 10t

→ 10/10 = t

→ 1 = t

→ t = 1 second

→ Time = 1 second