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.
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.
3 answers
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
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.
1 answer