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.
vi= 18 m/sec constant
vj= 24-gt= 24-9.8*6= 24-58.8= -34.8 m/sec
v^2= vx^2+vy^2= 18^2+(-34.8)^2=1535.04
KE= 1/2mv^2
KE= .5*1*1535.8= 767.9 joules
The answer depends upon which unit vector is "up". I assumed i; Richard assumed j.
To calculate the components of velocity at t = 6s, we can use the equations of motion.
a) The equation for vertical velocity (Vy) when an object is thrown upwards is:
Vy = Vyo - gt
Where:
Vyo = initial vertical velocity
g = acceleration due to gravity (approximated as 9.8 m/s^2)
Given that the initial vertical velocity is 24 m/s and the time is 6 seconds, we can substitute these values into the equation to find the vertical component of velocity at t = 6s:
Vy = 24 - (9.8 * 6)
Vy = 24 - 58.8
Vy = -34.8 m/s
The negative sign indicates that the ball is moving downwards.
b) To calculate the horizontal component of velocity (Vx), we know that there is no acceleration acting in the horizontal direction, so the initial horizontal velocity remains constant throughout the motion. Therefore, Vx = Vxo = 18 m/s.
To find the components of the velocity at t = 6s:
V = √(Vx^2 + Vy^2)
Substituting the values we obtained from parts a and b:
V = √(18^2 + (-34.8)^2)
V ≈ 39.92 m/s
Therefore, the components of the velocity at t = 6s are approximately Vx = 18 m/s and Vy = -34.8 m/s.
Now, let's move on to calculating the kinetic energy of the ball at t = 6s:
The kinetic energy (KE) of an object is given by the equation:
KE = (1/2) * m * V^2
Where:
m = mass of the object
V = magnitude of velocity
Given that the mass of the ball is 1 kg and the magnitude of velocity is approximately 39.92 m/s, we can substitute these values into the equation:
KE = (1/2) * 1 * (39.92)^2
KE ≈ 799.008 J
Therefore, the kinetic energy of the ball at t = 6s is approximately 799.008 Joules.
V(t) = (18 - gt)i + 24j
V(6) = -50.8 i + 24 j
KE (t=6) = (M/2)[50.8^2 + 24^2]
M = 1 kg
This assumes that the ball does not hit the ground before t = 6 seconds.