I throw a ball, mass m=5.0kg, at an initial speed of 3.0m/s, at an initial heigh
ID: 1917398 • Letter: I
Question
I throw a ball, mass m=5.0kg, at an initial speed of 3.0m/s, at an initial height of 1.7m, but at an angle of 37 degree with respect to the horizontal. What is the initial velocity? (This is a VECTOR. Define + i as the horizontal, and + j as the vertical.) What is the time when the ball obtains its maximum height? What is the maximum height of the ball? What is the velocity at the maximum height? It is NOT zero! What is the acceleration at the maximum height? What is the velocity JUST BEFORE it hits the ground? We define the Kinetic energy of an object as KE = KE = 1 / 2mv2. We define the Potential energy due to gravity as PE = mgh, where h is the HEIGHT of the object with respect to the ground. Obtain the Kinetic energy of the cannonball just when it left the cannon. Obtain the Potential energy of the cannonball just when it left the cannon. ADD the value of (g) and (h) together. Obtain the kinetic energy of the cannonball when it reaches the maximum height. Obtain the Potential energy of the cannonball when it reaches the maximum height. ADD the value of (j) and (k) together. Obtain the kinetic energy of the cannonball JUST before it hits the ground. Obtain the kinetic energy of the cannonball JUST before it hits the ground. ADD the value of (m) and (n) together. Look at your answers to (i), (1), and (o). EXPLAIN what's going on, in terms of the total energy of the system.Explanation / Answer
a)2.4i+1.805j
b)1.805/9.8 = 0.184s
c)1.866
d)2.4i
e)-9.8j
f)mgh + 0.5mu^2 (initial energy) = 0.5mv^2 final energy = > v = 6.5054 m/s