At time t=0,a 1kg ball is thrown up from the top of a tall tower with velocity Vo=18i+24j(m/s).
a:calculate the components of th velocity at t=6s.
b:Calculate the kinetic energy of the ball at t=6s.
To answer part (a), we need to consider the physics of projectile motion and apply the relevant equations. Let's break it down step by step:
Step 1: Split the velocity vector into its x-component and y-component.
Given: Vo = 18i + 24j (m/s)
The x-component of the velocity, Vx, remains constant throughout the motion. Therefore, Vx = 18 m/s.
The y-component of the velocity, Vy, changes due to the gravitational force acting on the ball.
Step 2: Calculate the final velocity in the y-direction using the equation Vy = Vo - gt, where g is the acceleration due to gravity (approximately -9.8 m/s^2).
At t = 6s, we have: Vy = 24 - (9.8 * 6) = -34.8 m/s
Step 3: The components of the velocity at t = 6s are:
Vx = 18 m/s (remains constant)
Vy = -34.8 m/s (negative sign indicates downward direction)
Now, moving on to part (b), we can calculate the kinetic energy of the ball at t = 6s. The kinetic energy (KE) is given by the equation KE = 1/2 * mv^2, where m is the mass of the ball and v is its velocity magnitude.
Given: Mass of the ball = 1 kg
Velocity at t = 6s: V = √(Vx^2 + Vy^2) = √(18^2 + (-34.8)^2) = 39.6 m/s
Using the above values, we can calculate the kinetic energy:
KE = 1/2 * 1 kg * (39.6 m/s)^2 = 784.08 J
Therefore, the kinetic energy of the ball at t = 6s is 784.08 Joules.