A hopper jumps straight up to a height of 1.7 m. With what velocity did it leave the floor? (acceleration = 10 m/s2)

Question

How do you arrive at -3.22 which is the answer I don't get it.

Answer 1

The kinematic equation you should use is: V(f) ^2 - V(i) ^2 = 2ah

Where V(f) = the final velocity

V(i) = the initial velocity

a = acceleration of gravity = -10

h = height that it rises

When the hopper reacher the top, the final velocity zero

So, plug in the rest of the values and

v(f)^2 =

(1.7)^2 = 0 2(10)h

h = 2.89/20 = 0. 145 m

Answer 2

David

v² = u² + 2as

If we assume direction upwards is positive

v = 0 (assuming instantaneous velocity is 0 when he gets to 1.7 m height)

a = - 10 m/s/s (minus because it is downwards)

s = 1.7

So 0 = u² - 2 x 10 x 1.7

u² = 34

u = 5.8 m/s (data given to 2 significant figures so quote result to the same)

Mike