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A ball is thrown vertically upwards from a height of 8 ft with an initial velocity of 30 ft per second. How high will the ball go? Note that the acceleration of the ball is given by a(t)= -32 feet per second per second
a) 50.1875 ft
b) 18.5469 ft
c) 53.1875 ft
d) 22.0625 ft
e) 34.1875 ft

Respuesta :

boffeemadrid

Answer:

c) [tex]22.0625\ \text{ft}[/tex]

Step-by-step explanation:

u = Initial velocity = 30 ft/s

s = Displacement

v = Final velocity = 0

a = Acceleration due to gravity = [tex]-32\ \text{ft/s}^2[/tex]

From kinematic equations we have

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-30^2}{2\times -32}\\\Rightarrow s=14.0625\ \text{ft}[/tex]

Distance the ball will reach from the height from which the ball is thrown is [tex]14.0625\ \text{ft}[/tex].

Distance of the ball from the ground is [tex]8+14.0625=22.0625\ \text{ft}[/tex].

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