A raindrop strikes the ground with a speed of 8.5 m/s. The specific heat of water is 4.19 x 103 J/(kg•°C). What is the raindrop’s change in temperature after it strikes the ground, in degrees Celsius, assuming all its initial kinetic energy goes into heating it?

0.00862 °C

Explanation:

When all Kinetic energy goes Into heating we can write

Q= mc × change in temperature

Q = 1/2 mv^2

1/ 2 mv^2 = mc × change in temperature

cancel out m because its in both sides

1/2 v^2 = c × change in temperature

1/ 2 × 8.5 ×8.5 = 4190 × x

x = 0.00862 °C

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west