A person is riding on a flatcar traveling at a constant speed v1= 15 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop. The hoop is at a height h=4 m above his hand. He throws the ball with a speed v2= 18 m/s with respect to the flatcar. Let g=10 m/s2 and neglect air drag completely.
(a) At what horizontal distance x in front of the hoop must the person release the ball? (in meters)
x=
(b) When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the flatcar? (angle αcar with respect to the horizontal in degrees)
αcar=
(c) When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the ground? (angle αground with respect to the horizontal in degrees)
Answers & Comments
Verified answer
(a)
It is probably easiest to imagine that the flatcar is stationary while the hoop approaches the person at 15 m/s. Even better, we can imagine that the hoop is stationary at the point at which the ball passes through it, 4 m above and x metres in front of the person.
If the vertical component of the velocity of the ball is initially U at the point of projection, and becomes V=0 when it passes through the hoop after travelling a vertical distance of h=4m, then
V^2=U^2-2gh
U^2=2gh=2*10*4=80 m2/s2
U=4sqrt(5) m/s.
The time taken to reach the maximum height is
t=U/g=4sqrt(5)/10=(2/5)sqrt(5) s.
If the horizontal component of the velocity of the ball relative to the flatcar is W (this is constant), then
U^2+W^2=(v2)^2
W^2=(v2)^2-U^2
=18^2-80
=244
W=2sqrt(61).
The horizontal distance travelled by the ball to the point of maximum height (ie where it passes through the hoop) is
x=Wt
=2sqrt(61)*(2/5)sqrt(5)
=(4/5)sqrt(5*61)
=14 m (to 2 sig figs, same as g).
(b)
tanA=U/W
=4sqrt(5)/2sqrt(61)
0.572598334
A=0.520027491rad
=30deg (to 2 sig figs).
(c)
As seen from the ground, the vertical component of velocity is the same, but the horizontal component is now W+v1. so
tanB=U/(W+v1)
=4sqrt(5)/(2sqrt(61)+15)
=0.292100785
B=0.284194146rad
=16deg (to 2 sig figs).
v2= 18 m/s, v1= 15 m/s, h = 4 m above his hand, g=10 m/s2
Let direction of velocity vector of the ball with respect to horizontal flatcar be (a)
Height y above hand at time t secs is given by
y(t) = [v2sin(a)]t - (1/2)gt^2 = t[18sin(a) - 5t] .............(1)
y is zero at start and again after t = (18/5)sin(a)
Neglecting air drag so assuming parabolic flight curve, by symmetry, top of curve
occurs after t(peak) = (9/5)sin(a) = (9s/5) for short ........(2)
a) The height above hand of top of curve is given by substituting this t into (1)
y(max) = (9s/5)[18s - 9s] = 81s^2/5 = 4 m
sin(a) = â(20/81)
cos(a) = â(61/81)
x(t) = (v1 + v2cos a)t so if we substitute t(peak) we can find release distance.
Person should release the ball at [15 + 18cos(a)]*t(peak) in advance of hoop.
That distance is [15 + 18*â(61/81)]*[(18/5)â(20/81)] ~ 42.943 metres
b) Angle a has sine of â(20/81) so a is ~ 29.795 degrees
c) From ground vertical component is the same, 18sin(a) = 2â(20) = 4â(5)
But horizontal component has added speed of flatcar, so it is
15 + 18cos(a) = 15 + 18â(61/81) = 15 + 2â(61)
From ground, new angle (b) is given by tan(b) = [4â(5)]/[15 + 2â(61)]
b ~ 16.28 degrees.
Regards - Ian