Question

Suppose the ball is thrown from the same height as in the PRACTICE IT(h = 41.0) problem at an angle of 28.0° below the horizontal. If it strikes the ground 47.6 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v₀t from the equation for the y-displacement.)

(a) the time of flight
s

(b) the initial speed
m/s

(c) the speed and angle of the velocity vector with respect to the horizontal at impact

speed	m/s
angle	° below the horizontal

Answer 1

(a) Suppose, horizontal component of velocity = Vox

And, vertical component of velocity = Voy

Time of flight = t

Now,

Vox*t = 47.6

t = 47.6 / Vox

again,

-41 = -Voy*t - g/2*t^2

41 = Voy*47.6/Vox + 4.9*47.6^2/Vox^2

Voy/Vox = tan 28°= 0.532

41 - 47.6*0.532 = 4.9*47.6^2/Vox^2

Vox = √4.9*47.6^2 / (41-47.6*0.532)

= √11102.224/15.6768

= 26.61 m/s

So,

Initial speed, Vo = Vox/cos 28 = 26.61/0.883 = 30.14 m/sec

Therefore,

Time of flight t = d/Vox = 47.6/26.61 = 1.79 sec (Answer)

(b) As we have calculated above, Initial speed, Vo = 30.14 m/sec (Answer)

(c) Final vertical speed Vy = -Voy-g*t = -30.14*sin 28 - 9.8*1.79

= -14.15 - 17.54

= - 31.69 m/sec

impact speed V = √Vy^2+Vox^2

= √[(-31.69)^2 + 26.61^2] = 41.38 m/sec (Answer)

impact angle Θi = arctan (Vy/Vox) = arctan(-31.69/26.61) = -50.0° = 50.0° (Below the horizontal)