Respuesta :
Answer:
The correct option is;
a. 20 m/s
Step-by-step explanation:
The given parameters are;
The angle at which the ball is thrown, θ = 30° to the horizontal
The horizontal distance of the top edge of the building where the ball lands from where the ball is thrown, x = 20 m
The height of the top edge of the building above the throwing point = 5 meters
Let "v" represent the speed with which the ball is thrown
We have;
The vertical component of the speed with which the ball is thrown, [tex]v_y[/tex] = v à sin(θ) = v à sin(30°) = v à 0.5 = 0.5¡v
[tex]v_y[/tex] = 0.5¡v
The horizontal component of the speed with which the ball is thrown, vâ = v Ă cos(θ) = v Ă cos(30°) = v Ă 0.9 = 0.9¡v
vâ = 0.9¡v
The kinematic equation of the motion is y = [tex]v_y[/tex]¡t - (1/2)¡g¡t², where;
y = The vertical height reached = 5 metes
t = The time taken to reach the specified 5 m, height
g = The acceleration due to gravity = 9.8 m/s², we have;
Therefore, we have;
5 = 0.5¡v¡t - (1/2)¡9.8¡t²...(1)
Also, from the horizontal motion of the ball, we have the following kinematic equation of motion;
x = vâ Ă t
Therefore, by substituting the known values, we have;
20 = 0.9¡v à t
ⴠv = 20/(0.9¡t) = 200/(9¡t)...(2)
Substituting the value of t in equation (1) gives;
5 = 0.5¡v¡t - (1/2)¡9.8¡t² = 0.5¡(200/(9¡t))¡t - (1/2)¡9.8¡t²
ⴠ5 = 0.5¡(200/(9¡t))¡t - (1/2)¡9.8¡t² = 100/9 - 4.9¡t²
4.9¡t² = 100/9 - 5 = 55/9
t = â(55/(9 Ă 4.9)) â 1.116766
The time taken to reach the specified 5 m height = t â 1.116766 seconds
From equation (2), we have, v = 200/(9¡t) = 200/(9 Ă 1.116766) â 19.8987 m/s
The speed with which the ball is thrown = v â 19.8987 m/s â 20 m/s. to the nearest whole number.
The speed with which the ball is thrown is approximately 20 m/s
