Question

A body
is thrown with the
velocity of 40m/s in direction making
an angle of 30 with the horizontal
(1) Horizontal range
[1] Maximum height
[1] Time taken to reach maximum height​

Answer 1

Answer:

Hey mate, here's your answer.

Explanation:

ANSWER

R.E.F image  

u=40m/s

θ=30  

∘

 

g=9.81

Time of flight, T=  

g

2usinθ

​  

 

Maximum Height, H=  

2g

u  

2

sin  

2

θ

​  

 

Horizontal Range,  R=  

g

u  

2

sin2θ

​  

 

Time taken to reach the  

max.height, t =  

2

T

​  

 

4=  

2×9.8

(40)  

2

(sin30  

∘

)  

2

 

​  

=20.408m

=20.41m

R=  

9.8

(40)  

2

sin2×30  

∘

 

​  

=141.4m

t=  

2

I

​  

=  

g

usin3θ

​  

=2.041s

Ans =141.4m,20.41m,2.041s

Answer 2

$\mathtt{ \huge{\fbox{Solution :)}}}$

Given ,

• Initial velocity (u) = 40 m/s
• Projection angle (Φ) = 30° degrees

We know that ,

$\large{ \mathtt{ \fbox{Horizontal \: range = \frac{ {u}^{2} \times \: \sin(2 \theta) }{g} }}}$

Thus ,

$\sf \mapsto Horizontal \: range = \frac{ {(40)}^{2} \times \sin((2 \times 30)) }{10} \\ \\ \sf \mapsto Horizontal \: range = \frac{1600 }{10 } \times \frac{ \sqrt{3} }{2} \\ \\ \sf \mapsto Horizontal \: range = 80 \sqrt{3} \: \: m$

Hence , the horizontal range is 80√3 m

$\bigstar \: \mathtt{\fbox{Maximum \: height = \frac{ {u}^{2} { \sin}^{2} \theta }{2g} }</p><p></p><p>}$

Thus ,

$\sf \mapsto Maximum \: height = \frac{ {(40)}^{2} \times { \sin}^{2}(30) }{2 \times 10} \\ \\ \sf \mapsto Maximum \: height = \frac{1600}{20} \times {( \frac{1}{2} )}^{2} \\ \\ \sf \mapsto Maximum \: height = \frac{80}{4} \\ \\ \sf \mapsto Maximum \: height = 20 \: \: m$

Hence , the maximum height is 20 m

$\bigstar \: \mathtt{\fbox{Time \: taken \: to \: reach \: maximum \: height = \frac{u \sin( \theta) }{g} }</p><p></p><p>}$

Thus ,

$\sf \mapsto Time \: taken = \frac{ 40 \times \sin( 30) }{10} \\ \\\sf \mapsto Time \: taken = 4 \times \sin(30) \\ \\\sf \mapsto </p><p>Time \: taken= 4 \times \frac{1}{2} \\ \\ \sf \mapsto Time \: taken = 2 \: \: sec$

Hence , the time taken to reach maximum height is 4 sec

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