A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in meter ,is_________
[tex] \colorbox{yellow}{\red{Correct \: answer \: only \: ✔}}[/tex]
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Verified answer
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[tex] \sf \: .H1 = \dfrac{ {u}^{2} { \sin }^{2}45 }{2g} = 120[/tex]
[tex] \sf \: \dfrac{ {u}^{2} }{4g} = 120...(i)[/tex]
When half of kinetic energy is lost
[tex] \sf \: v = \dfrac{u}{ \sqrt{2} } [/tex]
[tex] \sf \: H2 = \dfrac{( \dfrac{u}{ \sqrt{2} }) ^{2} {sin}^{2} 30 }{2g} = \dfrac{ { u}^{2} }{16g}...(ii)[/tex]
From (i) and (ii)
[tex] \sf \: H2 = \dfrac{H1}{4} [/tex]
[tex]{ \red{ \boxed{ \red{ \sf \:= 30 m }}}}[/tex]
[tex]{ \pink{ \sf \: hope \: it \: helps \: u}}[/tex]
Explanation:
The maximum height of the projectile is given by:
```
H = u^2 * sin^2 (45°) / 2g
```
where u is the initial velocity of the projectile and g is the acceleration due to gravity.
Substituting the given values, we get:
```
H = 120 m
```
Also, the kinetic energy of the projectile is given by:
```
KE = u^2 / 2
```
Since the kinetic energy is halved after the bounce, the new kinetic energy is:
```
KE = u^2 / 4
```
The velocity of the projectile after the bounce is given by:
```
v = u / 2
```
The maximum height that the projectile reaches after the bounce is given by:
```
H = v^2 * sin^2 (30°) / 2g
```
Substituting the given values, we get:
```
H = (u / 2)^2 * sin^2 (30°) / 2g
```
```
H = u^2 * sin^2 (30°) / 8g
```
```
H = 120 m / 8 = 15 m
```
Therefore, the maximum height that the projectile reaches after the bounce is **15 meters**.
**Correct The answer is 15 m.**