Respuesta :
Answer:
(A) Initial velocity of baseball is 14.97 m/s.
(B) Velocity of baseball just before hitting the wall is 11.16 m/s.
(C) The baseball is moving in downward direction just before striking the wall.
Explanation:
Let u m/s be the initial velocity of the baseball. According to the problem, the ball makes an angle of 55⁰ with respect to horizontal direction, Hence, the components of initial velocity along horizontal (u₁) and vertical direction (u₂) are:
u₁ = u cos 55⁰
u₂ = u sin 55⁰
The equation of motion is:
d = ut + 0.5at²
(A) Since baseball has no acceleration along horizontal direction, the horizontal distance covered by the baseball is:
d = u₁t
17 = u cos 55⁰ x t
29.64 = ut ......(1)
In vertical direction, acceleration due to gravity(g) is acted on the baseball, so the equation becomes:
h = u₂t - 0.5gt²
5 = u sin 55⁰ t - 0.5 x 9.8 x t² .......(2)
Substitute equation (1) in equation (2);
5 = 29.64 x 0.82 - 4.9 x t²
t = 1.98 s
Substitute the value of t in equation (1):
29.64 = u x 1.98
u = 14.97 m/s
(B) Let v be the final velocity of the baseball. The horizontal component of final velocity of baseball remains same as the initial velocity. Thus,
v₁ = u cos 55⁰ = 14.97 x cos 55⁰ = 8.58 m/s
The vertical component of final velocity of the baseball is:
v₂ = u sin 55⁰ - gt = 14.97 x sin 55⁰ - 9.8 x 1.98 = -7.14 m/s
The resultant final velocity of the baseball is:
v = [tex]\sqrt{v_{1} ^{2} + v_{2} ^{2} }[/tex]
v = [tex]\sqrt{8.58^{2} + (-7.14) ^{2} }[/tex]
v = 11.16 m/s
(C) The direction of the baseball just before hitting the wall is:
θ = tan⁻¹(v₂/v₁)
θ = tan⁻¹(-7.14/ 8.58)
θ = -39.76⁰
Thus, the baseball is moving in the downward direction.
(A) Initial velocity of baseball is 14.97 m/s.
(B) Velocity of baseball just before hitting the wall is 11.16 m/s.
(C) The baseball is moving in downward direction just before striking the wall.
Let u m/s be the initial velocity of the baseball.
It is given that the ball makes an angle of 55⁰ with respect to horizontal direction.
[tex]u_1 = u cos 55^o\\\\u_2 = u sin 55^o[/tex]
The equation of motion is:
[tex]d = ut + 0.5at^2[/tex]
(A) Since baseball has no acceleration along horizontal direction, the horizontal distance covered by the baseball is:
[tex]d = u_1t\\\\17 = u cos 55^o * t\\\\ut=29.64[/tex].................(1)
In vertical direction, acceleration due to gravity(g) is acted on the baseball, so the equation becomes:
[tex]h = u_2t - 0.5gt^2\\\\5 = u sin 55^o t - 0.5 * 9.8 * t^2[/tex] ......................(2)
On substituting equation (1) in equation (2);
[tex]5 = 29.64 * 0.82 - 4.9 * t^2\\\\t = 1.98 s[/tex]
Substitute the value of t in equation (1):
[tex]29.64 = u * 1.98\\\\u = 14.97 m/s[/tex]
(B) Let v be the final velocity of the baseball. The horizontal component of final velocity of baseball remains same as the initial velocity. Thus,
[tex]v_1 = u cos 55^o \\\\v_1= 14.97 * cos 55^o\\\\ v_1= 8.58 m/s[/tex]
The vertical component of final velocity of the baseball is:
[tex]v_2 = u sin 55^o - gt \\\\v_2= 14.97 * sin 55^o - 9.8 * 1.98\\\\ v_2= -7.14 m/s[/tex]
The resultant final velocity of the baseball is:
[tex]v=\sqrt{v_1^2+v_2^2} \\\\v=\sqrt{8.58^2+(-7.14)^2} \\\\v=11.16m/s[/tex]
(C) The direction of the baseball just before hitting the wall is:
[tex]\theta = tan^{-1}\frac{v_2}{v_1} \\\\\theta = tan^{-1}\frac{-7.14}{5.58}\\\\\theta = -39.76^o[/tex]
Thus, the baseball is moving in the downward direction.
Find more information about "Velocity" here:
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