Question

#15

Problem 3.28 《 15of16 Part A An athiete performing a long jump leaves the ground at a 30.2 angle and lands 7.62 m away Value Units Request Answer nger weld to rp be? this speed were neeased byMst 50% how much Ad-Value Units Provide Feedback MacBook Pro 4

Answer 1

1.

In projectile motion range of motion is given by:

R = Vx*T

Vx = initial speed in x-direction = V*cos A

V = initial speed of projectile at an angle of A with +ve x-axis

T = total time period of motion = 2*Vy/g = (2*V*sin A)/g

R = V*cos A*(2*V*sin A)/g = (V^2*sin 2A)/g

Now Using this equation for given situation:

R = (V^2*sin 2A)/g

V = sqrt (R*g/sin 2A)

Given that

R = 7.62 m

g = 9.81 m/sec^2

A = 30.2 deg

2A = 60.4 deg

So,

V = sqrt (7.62*9.81/sin 60.4 deg)

V = 9.27 m/sec = takeoff speed of athlete

Part B.

Speed is increased by 5.0%, then new takeoff speed will be

V1 = V + 5.0% of V = 1.05*V

V1 = 1.05*9.27 = 9.7335 m/sec = New takeoff speed

Now in this case range of athlete will be

R1 = (V1^2*sin 2A)/g

R1 = (9.7335^2*sin 60.4 deg)/9.81

R1 = 8.40 m

$\Delta d = R1 - R$

$\Delta d = 8.40 - 7.62 = 0.78 \; m$

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