Physics, asked by vrnpiyush, 1 year ago

A projectile takes off with an initial velocity of 10 m/s at an angle of elevation of 45°. It is just
able to clear two hurdles of height 2m each separated from each other by a distance d. Calculate
d. At what distance from the point of projection is the first hurdle placed? Take
g= 10 m/s2

Answers

Answered by JinKazama1
21

Answer:

 d = 2 \sqrt{5} m ,d_1 =( 5-  \sqrt{5}) m

Explanation:

We have,

u = 10m/s

 \theta =45\degree

h = 2 m

1) We know,

h_{max}=\frac{u^2sin^2(\theta) }{2g} \\ \\=\frac{10^2*sin^2(45\degree)}{2*10}=2.5m

2)

Time taken to reach from max. Height to top of hurdle:

u_y = 0 (at Max height)

S_y = 2.5-2=0.5m

Using,

S_y =u_yt+\frac{1}{2}gt^2 \\ \\=>0.5 = 0+5*t^2 \\ \\=>t =\frac{1}{\sqrt{10}} s

3) Now,

Time required to cross the hurdle from one to other is

 2t =\frac{2}{\sqrt{10}} s

Since, acceleration in horizontal direction is 0.

So,

 d = u_x*(2t) \\ \\ =10*cos(45\degree)*\frac{2}{\sqrt{10}}\\ \\= 2\sqrt{5}m

4) Range of Projectile,

 R = \frac{u^2sin(2\theta)}{g} \\ \\=\frac{10^2*sin(90\degree)}{10}\\ \\=10m

Then ,

Distance from point of projection to first hurdle placed is given by

 d_1 = \frac{R}{2}-\frac{d}{2} \\ \\ = (5-\sqrt{5})m

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