Patriciarecones97idn
Answered

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A ball was thrown vertically upwards with an initial velocity of 10 m/s. Suppose that the ball is positioned 1 m above the point of release, what will be the maximum height above the thrower's hand reached by the ball? How about the time it took the ball to reach its maximum height?​

Sagot :

Charish7955idn

Explanation:

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

#hope it helps :)

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