A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was lost in the process?
(Ciron = 0.444 J/g°C)
Q =mcAT

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J

The negative sign means heat is lost, which agrees with the decrease in temperature

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1