A ball is projected with an initial velocity upward component of 30 m/s and horizontal component of 23 m/s.
a. Find the velocity of the ball after 2 seconds
b. Find the position of the ball after 2 seconds
c. How long a time is required to reach the highest point of projectory?
d. How high is the point?
e. How long a time is required for the ball to return to its original level?
f. How far has it travelled horizontally during this time?
To solve these questions, we will break down the motion into its vertical and horizontal components. Let's start with the given information:
Vertical Component:
Initial vertical velocity (u_y) = 30 m/s
Vertical acceleration (a_y) = -9.8 m/s^2 (assuming acceleration due to gravity)
Horizontal Component:
Initial horizontal velocity (u_x) = 23 m/s
Horizontal acceleration (a_x) = 0 m/s^2 (assuming no external forces acting horizontally)
a. Find the velocity of the ball after 2 seconds:
To find the velocity of the ball after 2 seconds, we'll only consider the vertical component, as the horizontal component remains constant. We can use the formula for velocity in vertical motion:
v_y = u_y + a_y * t
Substituting the given values:
v_y = 30 m/s + (-9.8 m/s^2) * 2 s
v_y = 30 m/s - 19.6 m/s
So, the velocity of the ball after 2 seconds is v_y = 10.4 m/s (upward).
b. Find the position of the ball after 2 seconds:
To find the position of the ball after 2 seconds, we need to consider both the vertical and horizontal components separately.
Vertical motion:
Using the equation for displacement in vertical motion:
s_y = u_y * t + (1/2) * a_y * t^2
Substituting the given values:
s_y = 30 m/s * 2 s + (1/2) * (-9.8 m/s^2) * (2 s)^2
s_y = 60 m - 19.6 m
So, the vertical position of the ball after 2 seconds is s_y = 40.4 m (above the starting point).
Horizontal motion:
Since the horizontal component remains constant,
s_x = u_x * t
s_x = 23 m/s * 2 s
s_x = 46 m
So, the horizontal position of the ball after 2 seconds is s_x = 46 m.
c. How long a time is required to reach the highest point of trajectory?
The highest point of the projectile occurs when the vertical velocity becomes zero. To find the time it takes to reach this point, we'll use the equation:
v_y = u_y + a_y * t
Since the final vertical velocity is zero at the highest point, we have:
0 = 30 m/s + (-9.8 m/s^2) * t
Solving for t:
9.8 m/s^2 * t = 30 m/s
t = 30 m/s / 9.8 m/s^2
So, the time required to reach the highest point of the trajectory is t ≈ 3.06 seconds.
d. How high is the highest point?
To find the height of the highest point, we can use the equation for displacement in vertical motion:
s_y = u_y * t + (1/2) * a_y * t^2
Substituting the known values:
s_y = 30 m/s * 3.06 s + (1/2) * (-9.8 m/s^2) * (3.06 s)^2
So, the height of the highest point of the projectile is s_y ≈ 45.42 m.
e. How long a time is required for the ball to return to its original level?
The total time taken for the ball to return to its original level is twice the time it takes to reach the highest point, since the motion is symmetric. Therefore, the time required for the ball to return to its original point is approximately t ≈ 2 * 3.06 seconds.
f. How far has it traveled horizontally during this time?
The horizontal distance traveled during the entire motion is given by:
s_x = u_x * t
Substituting the known values:
s_x = 23 m/s * (2 * 3.06 s)
So, the horizontal distance traveled during this time is s_x ≈ 141.72 m.