Question

A ball is thrown upward with an initial velocity of 10 m/s. How high will it reach?
a. 1,02 m
b. 5,10 m
c. 0,51 m
d. 10,20 m

ayuda porfavor

Answer 1

Respuesta:

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 10m/s, v = 0 (at top), a = -g = -9.81m/s^2 and we want to find s. So, we use equation (2)

v^2 = u^2 + 2as

0 = 10^2 - 2(9.81)s

s = 100/19.62 = 5.1m

now to find t, we use equation (3)

v = u + at

0 = 10 – 9.81t

t = 10/9.81 = 1.02s

So, the ball reaches a height of 5.1m and it takes 1.02s to reach that height.

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