Question

an object is projected with velocity of 60m/s in a direction making an angle of 60° with the horizontal find the maximum height, the time taken to reach maximum height, the horizontal range​

Answer 1

Answer:

• Velocity (u) = 60 m/s
• Angle of projection = 60°
• Maximum height 'H' = ?
• Time taken to reach maximum height i.e time of ascent 'T' = ?
• Horizontal Range 'R' = ?

Maximum Height :

$\longrightarrow\red{\rm Height_{(maximum)} = \dfrac{u^2 \sin^2(\theta)}{2g}}$

$\longrightarrow\red{\rm Height_{(maximum)} = \dfrac{(60)^2 \sin^2( {60}^{ \circ} )}{2 \times 10}}$

$\longrightarrow\red{\rm Height_{(maximum)} = \dfrac{3600 \times \bigg(\frac{ \sqrt{3} }{2} \bigg)^{2} }{20}}$

$\longrightarrow\red{\rm Height_{(maximum)} = \dfrac{360 \times \frac{3}{4} }{2}}$

$\longrightarrow\red{\rm Height_{(maximum)} = \dfrac{\frac{1080}{4} }{2}}$

$\longrightarrow\red{\rm Height_{(maximum)} = \dfrac{270}{2}}$

$\longrightarrow \underline{ \underline{\red{\rm Height_{(maximum)} = 135 \: m} }}$

Time taken to reach maximum height i.e time of ascent :

$\longrightarrow\green{\rm Time_{(ascent)} = \dfrac{u \sin (\theta)}{g}}$

$\longrightarrow\green{\rm Time_{(ascent)} = \dfrac{60 \sin ( {60}^{ \circ} )}{10}}$

$\longrightarrow\green{\rm Time_{(ascent)} = \dfrac{60 \times \frac{ \sqrt{3} }{2} }{10}}$

$\longrightarrow\green{\rm Time_{(ascent)} = \dfrac{6 \sqrt{3} }{2} }$

$\longrightarrow\green{\rm Time_{(ascent)} =3 \sqrt{3} }$

$\longrightarrow\green{\rm Time_{(ascent)} =3 \times 1.73}$

$\longrightarrow \underline{ \underline{\green{\rm Time_{(ascent)} =5.19 \: s}}}$

Horizontal Range :

$\longrightarrow\pink{\rm Range_{(Horizontal)} = \dfrac{u^2 \sin2(\theta)}{g}}$

$\longrightarrow\pink{\rm Range_{(Horizontal)} = \dfrac{(60)^2 \sin2( {60}^{ \circ} )}{10}}$

$\longrightarrow\pink{\rm Range_{(Horizontal)} = \dfrac{3600 \sin( {120}^{ \circ} )}{10}}$

$\longrightarrow\pink{\rm Range_{(Horizontal)} = 360 \sin( {120}^{ \circ} )}$

$\longrightarrow\pink{\rm Range_{(Horizontal)} = 360 \times \dfrac{ \sqrt{3} }{2} }$

$\longrightarrow\pink{\rm Range_{(Horizontal)} = 180 \times \sqrt{3} }$

$\longrightarrow\pink{\rm Range_{(Horizontal)} = 180 \times 1.73 }$

$\longrightarrow \underline{ \underline{\pink{\rm Range_{(Horizontal)} = 311.4 \: m }}}$