Question

From ground to ground projectile at maximum height find range if height is 4 m also find time of flight.

Answer 1

Answer:

Range = 16 m

Time of flight = 1.503 sec

Explanation:

Given :  

Height ( H ) = 4 m

We know that  angle is maximum at 45

So ,  θ = 45

First let us find Range ( R )

We know ratio of Range and height  

$\displaystyle \frac{R}{H}= \frac{4}{\tan\theta}$

Now put the value here

$\displaystyle \frac{R}{4}= \frac{4}{\tan45} \\\\\displaystyle \text{ \tan45=1 So we get}\\\\\displaystyle R=16 \ m$

We also know formula for Range

$\displaystyle R=\frac{u^2\sin2\theta}{g}$

Put the value here to get u ( initial velocity )

Taking g = 10 $m/sec^2$

$\displaystyle \text{ 16=\frac{u^2\times2\times\frac{1}{\sqrt2}}{10} }\\\\\displaystyle 160=u^2\times\sqrt2\\\\\displaystyle \text{ u=\sqrt{\frac{160}{\sqrt2}} }\\\\\displaystyle u=\sqrt{80\sqrt2} \ m/sec\\\\\displaystyle u=10.63 \ m/sec$

Now for time of flight ( T )  we have

$\displaystyle T_f=\frac{2u\sin\theta}{g}$

Now put the value here

$\displaystyle T_f=\frac{2\times10.63\times\sin45}{10}\\\\\displaystyle T_f=\frac{\sqrt2\times10.63 }{10}\\\\\displaystyle T_f=\frac{ 15.03}{10}\\\\\displaystyle T_f=1.503 \ sec$

Thus we get answer.

Answer 2

» From ground to ground projectile at maximum height is 4 m.

Here ..

Height (H) = 4 m

_________ [ GIVEN ]

• We have to find Range (R) and time of flight (T)

____________________________

We know that

R = $\dfrac{ {u}^{2}sin2 \theta }{g}$

→ sin2Ø = 1

→ sin2Ø = sin 90°

→ Ø = 45° (max.)

_____________________________

Now..

R = 4H tanØ

Here .. H = 4 m and Ø = 45°

→ R = 4(4) tan45°

→ R = 16(1)

→ R = 16 m

_____________________________

» For Horizontal Range ..

R = $\dfrac{ {u}^{2}sin2 \theta }{g}$

Here ..

R = 16, Ø = 45° (sin 45° = 1/√2), g = 10

u = ?

→ 16 = $\dfrac{ {u}^{2} \: \times \: 2 \: \times \: \frac{1}{ \sqrt{2} } }{10}$

→ u² = $\dfrac{160\sqrt{2}}{2}$

→ u² = 80√2

→ u = $\sqrt{80 \sqrt{2} }$ m/s or 10.62 m/s

____________________________

» Time of flight :

T = $\dfrac{2u\:sin\theta}{g}$

Here ..

T = ?

u = 10.62, g = 10, sin45 = √2

→ T = $\dfrac{2\:\times\:10.62\:\times\:sin45}{10}$

→ T = $\dfrac{10.62\:\times\: \sqrt{2} }{5}$

→ T = 1.502 sec

______________________________

Range = 16 m

Time of flight = 1.5 sec (approx.)

________ [ ANSWER ]

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