Question 1 Procedure: Keep the initial velocity fixed at 50 m/s. Perform the exp

Question

Question 1 Procedure: Keep the initial velocity fixed at 50 m/s. Perform the experiment for each of the following angles: 15 degrees 30 degrees 45 degrees 60 degrees 75 degrees Record in table 1: The maximum horizontal distance (Range) traveled by the projectile at various angles The total time of flight at various angles The maximum height attained Table 1: Experiment 1 - Range & Time of flight S. No. Initial Velocity Angle of Projection Range Time of Flight Maximum Height Attained 1. 2. 3. 4. 5. Observe and Analyze: What angle produces the maximum range? Why? What angle produces the maximum height? Why? Are there angles which produce the same range? If so, how would you explain this? Question 2 Procedure: Set the angle to 45 degrees. Perform the experiment with the following initial velocities. 30 m/s 40 m/s 50 m/s 60 m/s Record in Table 2: The horizontal and vertical components of the velocity and the ranges for each of the velocities. Table 2: Experiment 2 - Range at 45

Explanation / Answer

Question 1.

range= v^2sin2(theta) /g

R(15)=50^2*sin(30)/10 = 125m

R(30)=50^2*sin(60)/10 = 216.50m

R(45)=50^2*sin(90)/10 = 250m

R(60)=50^2*sin(120)/10 = 216.50m

R(75)=50^2*sin(150)/10 = 125m

Range at 45 degree will be maximum as sin2theta = 1 ... that is maximum.

R(15)=R(75) ..... as sin30=sin150

R(30)=R(60) ..... as sin60=sin120

T= 2*v*sin(theta)/g

T(15 degree) = 2*50*sin(15 degree)/10 =2.58 s

T(30 degree) = 2*50*sin(30 degree)/10 = 5 s

T(45 degree) = 2*50*sin(45 degree)/10 = 7.07 s

T(60 degree) = 2*50*sin(60 degree)/10 = 8.66 s

T(75 degree) = 2*50*sin(75 degree)/10 = 9.65 s

T is increses as angle increses bcoz sin(theta) increses

max. height H= (vsin(theta))^2/2g

H(15 degree) = (50*sin(15 degree))^2/20 = 8.37m

H(30 degree) = (50*sin(30 degree))^2/20 = 31.25m

H(45 degree) = (50*sin(45 degree))^2/20 = 62.5m

H(60 degree) = (50*sin(60 degree))^2/20 = 93.75m

H(75 degree) = (50*sin(75 degree))^2/20 = 116.62m

H is increses as angle increses bcoz sin(theta) increses

question 2.)

V(30)x = V(30)y = 30*cos45 = 21.21

V(40)x = V(40)y = 40*cos45 = 28.28

V(50)x = V(50)y = 50*cos45 = 35.35

V(60)x = V(60)y = 60*cos45 = 42.42

R=range= v^2sin2(theta) /g

theta is 45 , so sin2(thet) = sin 90 = 1

R(v=30) = (30^2)/10 = 90 m

R(v=40) = (40^2)/10 = 160m

R(v=50) = (50^2)/10 = 250m

R(v=60) = (60^2)/10 = 360m

as R is directely proportional to sqare of v

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