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Grade: 11

                        

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

4 years ago

Answers : (5)

Tapas Khanda
27 Points
							
Let the angle of throwing be Q with respect to the ground
So, the range is R = 30*cosQ*t
where 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))/2
Now, using the equation of motion for the flight from the ground to top of its flight
v = u + at
For the vertical motion,
v = final vertical speed at top = 0 
u = initial vertical speed at ground = 40*sinQ
a = -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*R
So, 20*30*sin2Q = 4.9*R ←---- using the property 2*sinQ*cosQ = sin2Q
So, R = (600/4.9)*sin2Q = 122.4*sin2Q
So, Now, sine is an increasing function with maximum value at 90 deg such that sin2Q = 1
So, maximum value of R = 122.4 m 
4 years ago
Tapas Khanda
27 Points
							 
Let the angle of throwing be Q with respect to the ground
So, the range is R = 40*cosQ*t
where 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))/2
Now, using the equation of motion for the flight from the ground to top of its flight
v = u + at
For the vertical motion,
v = final vertical speed at top = 0 
u = initial vertical speed at ground = 40*sinQ
a = -9.8 m/s2
t = t/2
So, 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*R
So, 20*40*sin2Q = 4.9*R ←---- using the property 2*sinQ*cosQ = sin2Q
So, R = (800/4.9)*sin2Q = 163.3*sin2Q
Now, using the equation of motion,
v^2 = u^2 +2as
For vertical flight from ground to top,
v = final vertical speed = 0
u = 40*sinQ
s = vertical displacement = 25 m
So, 0^2 = (40*sinQ)^2 – 2*9.8*25
So, sinQ = 0.5534
So, Q = 33.6 deg
So, range, R =  163.3*sin(2*33.6 deg) = 150.5 m
4 years ago
Mayank kumar
12 Points
							Plz provide rhe diagram of this situation? How you find the valve tignometric angle is there not any easy way
						
4 years ago
Tapas Khanda
27 Points
							
You can directly use the formula to find range : R = (v^2/g)*sin2Q
Maximum 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 deg
So, R =( 40^2/9.8)*sin(2*33.6 deg)
= 150.5 m
4 years ago
Tapas Khanda
27 Points
							
Diagram is not required to solve this question. 
 
You can directly use the formula to find range : R = (v^2/g)*sin2Q
Maximum 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 deg
So, R =( 40^2/9.8)*sin(2*33.6 deg)
4 years ago
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