Respuesta :
Answer
given,
initial velocity of the base ball = 12.5 m/s
height of the building = 11.5 m
angle of launch = −37°
speed when it reaches ground = ?
using kinematics equation
velocity in y direction
[tex]v_y^2 = (v sin \theta)^2 + 2 gh[/tex]
[tex]v_y^2 = (12.5 sin 37^0)^2+ 2\times 9.8 \times 11.5 [/tex]
[tex]v_y^2 = (12.5 sin 37^0)^2+ 2\times 9.8 \times 11.5 [/tex]
v_y = 16.79 m/s
velocity in x- direction
v_x = v cos θ
v_x = 12.5 x cos 37°
v_x = 9.98 m/s
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
[tex]v = \sqrt{9.98^2 + 16.79^2}[/tex]
v = 19.53 m/s
b) if it is launched vertically
initial vertical velocity = 0
[tex]v_y = \sqrt{2gh}[/tex]
[tex]v_y = \sqrt{2\times 9.8 \times 11.5}[/tex]
v_y = 15 m/s
v_x = 12.5 m/s
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
[tex]v = \sqrt{12.5^2 + 15^2}[/tex]
v = 19.52 m/s
The speeds when the baseball reaches the ground in each instance given is;
1) v_f = 18.4 m/s
2) v_f = 19.54 m/s
1) We want to find the speed when the baseball reaches the ground if the launch angle is -37°.
Since initial velocity is 12.5 m/s, then vertical component of initial velocity is;
u = 12.5 sin 37
u_y = 7.523 m/s
Using the equation;
v² = u² + 2gs
at height of 11.5m, we have;
v² = 7.523² + 2(9.8 × 11.5)
v² = 282
v = √282
v_x = 16.79 m/s
Thus, speed when it reaches the ground is;
v_f = √(7.523² + 16.79²)
v_f = 18.4 m/s
2) If it is launched horizontally;
u_y = 0 m/s
s = 11.5 m
Thus;
v² = 0² + 2(9.8 × 11.5)
v² = 225.4
v = 15.01 m/s
Thus, speed when it reaches the ground is;
v_f = √(12.5² + 15.01²)
v_f = 19.54 m/s
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