Question

A body is thrown with a velocity of 40m/s at an angle of 30* above the horizontal calculate the time taken to reach the maximum height 2m

Answer 1

Given :-

▪ A body is thrown with a velocity of u = 40 m/s at an angle of 30° above the horizontal.

To Find :-

▪ Time taken to reach the Maximum height

Solution :-

We can find the time taken by using the time of flight formula of a free fall body.

⇒ T = u / g

But, we need the vertical component.

⇒ Vert. comp. = u sin θ

⇒ Vert. comp. = 40 × sin 30°

⇒ Vert. comp. = 40 × 1/2

⇒ Vert. comp. = 20 m/s

Let us use the time of flight formula now,

⇒ T = u / g

⇒ T = Vert. comp. / g

⇒ T = 20 / 10

⇒ T = 2 seconds.

Hence, The body would take 2 seconds to reach the maximum height.

Other Formulae :-

☞ Range = u² sin 2θ / g

☞ Maximum height = u² sin² θ / 2g

☞ Time of flight = 2u sinθ / g

☞ The range of a projectile is maximum when thrown at an angle of 45° with the horizontal.

*Important*

⇒ Range / Height = 4 / tan θ

• θ = angle with the horizontal.

Answer 2

GIVEN :-

• A ʙᴏᴅʏ ɪs ᴛʜʀᴏᴡɴ ᴡɪᴛʜ ᴀ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ 40ᴍ/s .

• Tʜᴇ ʙᴏᴅʏ ɪs ᴛʜʀᴏᴡɴ ᴀᴛ ᴀɴ ᴀɴɢʟᴇ ᴏғ 30° ᴀʙᴏᴠᴇ ᴛʜᴇ ʜᴏʀɪᴢᴏɴᴛᴀʟ .

• Hᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴘʀᴏᴊᴇᴄᴛɪʟᴇ ᴍᴏᴛɪᴏɴ ɪs "2ᴍ" .

TO FIND :-

• Tʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ᴛᴏ ʀᴇᴀᴄʜ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ .

SOLUTION :-

Wᴇ ʜᴀᴠᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\purple\checkmark\:\bf\pink{Time\:taken\:to\:reach\:the\:maximum\:height\:\atop{is\:the\:half\:of\:the\:time\:of\:flight\:,i.e.\:T/2\:.}\:}$

$\red\checkmark\:\bf\purple{Time\:taken\:to\:reach\:the\:max^{m}\:height=\:\dfrac{u\:\sin{\theta}}{g}\:}$

Wʜᴇʀᴇ,

• $\bf\red{u}$ = 40m/s

• $\bf\red{\theta}$ = 30°

• $\bf\red{g}$ = 10 m/s²

➳ Tɪᴍᴇ ᴛᴀᴋᴇɴ = $\rm{\dfrac{40\times{\sin{30^{\degree}}}}{10}\:}$

➳ Tɪᴍᴇ ᴛᴀᴋᴇɴ = $\rm{4\times{\dfrac{1}{2}}\:}$

➳ Tɪᴍᴇ ᴛᴀᴋᴇɴ = $\bf\green{2\:second}$

$\huge\red\therefore$ Tʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ᴛᴏ ʀᴇᴀᴄʜ ᴛʜᴇ ᴍᴀxɪᴍᴜᴍ ʜᴇɪɢʜᴛ is "2 second" .