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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

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

Grade:11

5 Answers

Tapas Khanda
27 Points
7 years ago
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 
Tapas Khanda
27 Points
7 years ago
 
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
Mayank kumar
12 Points
7 years ago
Plz provide rhe diagram of this situation? How you find the valve tignometric angle is there not any easy way
Tapas Khanda
27 Points
7 years ago
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
Tapas Khanda
27 Points
7 years ago
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)

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