Question

A shell is fired from a canon with a velocity of 200m/s at an angle of 30°above the horizontal. find its
1) maximum hight.
2) time to flight by the shell.​

Answer 1

Answer:

Given:

Velocity = 200 m/s

Angle.of projectile = 30°

To find :

Max height and time taken

Formulas used:

Max height = u²sin²(@)/2g

Time = 2u sin(@)/g,

where @ is the angle of the projectile, u is the initial velocity, g is gravity .

Calculation:

Max height = 200² sin²(30°)/20

=> Max height = 40000/(40 ×2)

=> Max height = 0.5 km.

Time = 2 × 200 × sin(30°)/10

=> time = 20 seconds.

Answer 2

Question :-

→ Given above ↑

Answer :-

(1) Maximum height= 500 m or 0.5 km

(2) Time of flight = 20 seconds

To Find :-

(1) Maximum height

(2) time of flight

Formula used :-

$\rightarrow \boxed{ \sf{h_{max.} = \frac{ {u}^{2} \sin \: \theta }{2g}} } \\ \\ \rightarrow \boxed{\sf{time \: of \: flight = \frac{2u \: \sin \: \theta}{g} }}$

Solution :-

Given that ,

$\star \: \: \sf{ velocity \: (u) = 200 \: \frac{m}{s} } \\ \\ \star \: \: \sf{angle \: ( \theta) = 30 \degree}$

Therefore ,

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(1) For maximum height ,

$\rightarrow \sf{ h_{max.} \: = \frac{200 \times 200 \: { \sin}^{2} \:30 \degree}{2 \times 10} } \: \because \: g = 10 \: \frac{m}{ {s}^{2} } \\ \\ \rightarrow \sf{h_{max.} \: = \frac{200 \times 200}{20 \times 4} } \: \\ \\ \rightarrow \sf{ h_{max.} \: = \frac{2000}{4} } \\ \\ \rightarrow \boxed{ \sf{ h_{max.} \: = 500 \: m}}$

Maximum height is 500 m.

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(2) For time of flight

$\rightarrow \sf{t \: = \frac{2 \times 200 \: \sin \: 30 \degree}{10} } \\ \\ \rightarrow \sf{ t \: = 40 \times \frac{1}{2} } \\ \\ \rightarrow \boxed{\sf{t \: = 20 \: seconds}}$

Hence time of flight is 20 seconds .

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