Question

A ball is thrown vertically upward with a speed of 25.0 m/s. (a) How high does it rise?
(b) How long does it take to reach its highest point? (c) How long does the ball take to
hit the ground after it reaches its highest point? (d) What is its velocity when it returns to
the level from which it started?

Answer 1

Answer:

Given Information:

• Initial Velocity (u) = 25 m/s

(a) Using the third equation of motion we get:

⇒ v² - u² = 2as

⇒ 0² - 25² = 2 ( -10 m/s² ) s

(Since at the highest point velocity (v) is 0 and opposite to earth's gravity implies acceleration is negative.)

⇒ -625 = -20 s

⇒ s = 625 / 20

⇒ s = 31.25 m

Hence it rises upto a height of 31.25 m.

(b) Using first equation of motion, we get:

⇒ v = u + at

⇒ 0 = 25 + (-10 m/s²) t

⇒ 0 - 25 = -10 t

⇒ 25 = 10 t

⇒ t = 25/10 = 2.5 seconds

Hence the ball takes 2.5 seconds to reach the highest point (31.25 m).

(c) After reaching the highest point,

• Initial velocity (u') = 0
• Distance (s') = 31.25 m
• Acceleration (a') = + 10 m/s²
• Time taken during descent (t') = ?

Using Second equation of motion we get:

⇒ s' = u'(t') + 0.5 a'(t')²

⇒ 31.25 = 0 + 0.5 × 10 × (t')²

⇒ 31.25 = 5(t')²

⇒ (t')² = 31.25/5 = 6.25

⇒ t' = √6.25 = 2.5 seconds

The ball takes 2.5 seconds to hit the ground after it reaches the highest point.

(d) Final Velocity at the point where it started can be calculated as:

⇒ v' = u' + a't'

⇒ v' = 0 + 10 × 2.5

⇒ v' = 25 m/s

Hence it has a velocity of 25 m/s at the point where it started.

Answer 2

$\large\underline{ \underline{ \text{Question:}}}$

A ball is thrown vertically upward with a speed of 25.0 m/s.

• (a) How high does it rise?

• (b) How long does it take to reach its highest point?

• (c) How long does the ball take to hit the ground after it reaches its highest point?

• (d) What is its velocity when it returns to the level from which it started?

.

$\large\underline{ \underline{ \text{Solution:}}}$

• $\text{Initial velocity }(u) = 2 {ms}^{ - 1} \: \: \: ... \text{given}$

(a) How high does it rise?

As we know that,

• The velocity at highest point is zero.

Because, Kinetic energy is zero that time.

So,

• $\text{Final velocity } (v)= 0{ms}^{ - 1}$

Let's take,

• $\text{Acceleration due to gravity} \: (g) = 10m {s}^{ - 2}$

As it is retarding.

So, Acceleration will negative.

• $a = - 10m {s}^{ - 2}$

According to 3rd law of motion,

$\longrightarrow \boxed{ {v}^{2} - {u}^{2} = 2as}$

We get,

$\longrightarrow {(0m {s}^{ - 1} )}^{2} - {(25m {s}^{ - 1} })^{2} = 2( - 10m {s}^{ - 2}) s$

$\longrightarrow {0 {m}^{2} {s}^{ - 2}} - {625 {m}^{2} {s}^{ - 2} } = - 20m {s}^{ - 2}s$

$\longrightarrow { - 625 {m}^{2} {s}^{ - 2} } = - 20m {s}^{ - 2}s$

$\longrightarrow \frac{ - 625 {m}^{2} {s}^{ - 2} }{- 20m {{s}^{ - 2}}} = s$

$\longrightarrow \boxed{s = 31.25m}$

(b) How long does it take to reach its highest point?

According to 1st law of motion,

$\longrightarrow \boxed{v = u + at }$

We get,

$\longrightarrow 0m {s}^{ - 1} = 25m {s}^{ - 1} + ( - 10m {s}^{ - 2}) t$

$\longrightarrow - 25m {s}^{ - 1} = - 10m {s}^{ - 2} t$

$\longrightarrow \frac{- 25m {s}^{ - 1} }{ - 10m {s}^{ - 2}} = t$

$\longrightarrow \boxed{t = 2.5s}$

(c) How long does the ball take to hit the ground after it reaches its highest point?

At the highest point,

• $\text{Initial velocity }(u_2) = 0 {ms}^{ - 1}$
• $\text{Distance} \: (s)= 31.25m$

As the ball is falling down.

The speed will increase.

So, Acceleration will be positive.

• $\text{Acceleration} \: (a) = 10m {s}^{ - 2}$

According to 2nd law of motion,

$\longrightarrow \boxed{s = u_2t + \frac{1}{2} a {t}^{2} }$

We get,

$\longrightarrow 31.25m = (0m {s}^{ - 1} )t + \frac{1}{2}(10m {s}^{ - 2}) {t}^{2}$

$\longrightarrow 31.25m = (5m {s}^{ - 2}) {t}^{2}$

$\longrightarrow \frac{31.25m}{ 5m {s}^{ - 2} } = {t}^{2}$

$\longrightarrow 6.25 {s}^{2} = {t}^{2}$

$\longrightarrow {(2.5s)}^{2} = {t}^{2}$

$\longrightarrow \boxed{t = 2.5s}$

(d) What is its velocity when it returns to the level from which it started?

According to 1st law of motion,

$\longrightarrow \boxed{v_{2} = u_{2} + at }$

We get,

$\longrightarrow v_{2} = 0 {ms}^{ - 1} + (10m {s}^{ - 2})(2.5s)$

$\longrightarrow v_{2} = 0 + (10m {s}^{ - 2})(2.5s)$

$\longrightarrow \boxed{ v_{2} = 25 {ms}^{ - 1}}$

$\large\underline{ \underline{ \text{Final Answer:}}}$

• (a) The ball will rise 31.25m from the ground level.

• (b) It takes 2.5s to reach highest point.

• (c) It takes 2.5s to reach the ground.

• (d) The final velocity of ball while returning to the ground is 25m/s.