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The cieling of long hall is 25 m high What is the maximum horizontal distance that a ball thrown with a speed of 40 ms can go without hitting the ceiling of wall? Plz explain with a DIAGRAM

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4 years ago

```							Let the angle of throwing be Q with respect to the groundSo, the range is R = 30*cosQ*twhere t = time taken reach ground.Now, we know the time taken to reach the top of its flight is half the total time taken to reach the ground,So, time taken to reach the top of its flight = t/2 = (R/(30*cosQ))/2Now, using the equation of motion for the flight from the ground to top of its flightv = u + atFor the vertical motion,v = final vertical speed at top = 0 u = initial vertical speed at ground = 40*sinQa = -9.8 m/s2 t = t/2 So, 0 = 40*sinQ – 9.8*(t/2) So, 40*sinQ = 4.9*t = 4.9*(R/(30*cosQ))So, 40*30*sinQ*cosQ = 4.9*RSo, 20*30*sin2Q = 4.9*R ←---- using the property 2*sinQ*cosQ = sin2QSo, R = (600/4.9)*sin2Q = 122.4*sin2QSo, Now, sine is an increasing function with maximum value at 90 deg such that sin2Q = 1So, maximum value of R = 122.4 m
```
4 years ago
```							 Let the angle of throwing be Q with respect to the groundSo, the range is R = 40*cosQ*twhere t = time taken reach ground.Now, we know the time taken to reach the top of its flight is half the total time taken to reach the ground,So, time taken to reach the top of its flight = t/2 = (R/(40*cosQ))/2Now, using the equation of motion for the flight from the ground to top of its flightv = u + atFor the vertical motion,v = final vertical speed at top = 0 u = initial vertical speed at ground = 40*sinQa = -9.8 m/s2t = t/2So, 0 = 40*sinQ – 9.8*(t/2) So, 40*sinQ = 4.9*t = 4.9*(R/(40*cosQ))So, 40*40*sinQ*cosQ = 4.9*RSo, 20*40*sin2Q = 4.9*R ←---- using the property 2*sinQ*cosQ = sin2QSo, R = (800/4.9)*sin2Q = 163.3*sin2QNow, using the equation of motion,v^2 = u^2 +2asFor vertical flight from ground to top,v = final vertical speed = 0u = 40*sinQs = vertical displacement = 25 m So, 0^2 = (40*sinQ)^2 – 2*9.8*25So, sinQ = 0.5534So, Q = 33.6 degSo, range, R =  163.3*sin(2*33.6 deg) = 150.5 m
```
4 years ago
```							Plz provide rhe diagram of this situation? How you find the valve tignometric angle is there not any easy way
```
4 years ago
```							You can directly use the formula to find range : R = (v^2/g)*sin2QMaximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8)So, sin^2Q = 0.3063 => Q = 33.6 degSo, R =( 40^2/9.8)*sin(2*33.6 deg)= 150.5 m
```
4 years ago
```							Diagram is not required to solve this question.  You can directly use the formula to find range : R = (v^2/g)*sin2QMaximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8)So, sin^2Q = 0.3063 => Q = 33.6 degSo, R =( 40^2/9.8)*sin(2*33.6 deg)
```
4 years ago
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