A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are: a) 10 m/s; 9.8 m/s2.
b) 9.0 m/s; 9.0 m/s2.
c) 9.0 m/s; 9.8 m/s2.
d) 1.0 m/s; 9.8 m/s2.
e) 1.0 m/s; 9.0 m/s2.

a) 10 m/s; 9.8 m/s²

Explanation:

After leaving the bench the boy undergoes a semi-projectile motion from top to bottom. Neglecting the effects of air friction, we can safely assume that the horizontal velocity remains constant throughout the motion. Thus, the magnitude of velocity is 10 m/s. The only acceleration in the boy is the acceleration due to gravity, due to free fall motion. Therefore, the magnitude of acceleration is 9.8 m/s².

Therefore, One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:

a) 10 m/s; 9.8 m/s²

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

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right