[ام عبد الرحمن محمد]

سؤال 14
How far horizontally from the base of the building does the ball strike the ground? >>

X = Vx(T)

where

X = horizontal distance from base of building where ball will land
Vx = horizontal component of velocity
T = time ball strikes the ground after being tossed

Substituting appropriate values,

X = 8(cos 20)(3)

X = 22.55 m


<< Fund the height from which the ball was thrown. >>

Y = Vy(T) + (1/2)(g)T^2

Y = height from which ball was thrown
Vy = vertical component of the velocity
T = time when ball strikes the ground
g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting appropriate values,

Y = 8(sin 20)(3) + (1/2)(9.8)(3^2)

Y = 52.31 meters


<< How long does it take the ball to reach a point 10.0 m below the level of launching? >>

Formula is

S = Vy(T) + (1/2)(g)T^2

where

S = 10 m
Vy = 8(sin 20)
g = 9.8 m/sec^2
T = time for ball to reach a point 10 m below the launching level

Substituting values,

10 = 8(sin 20)T + (1/2)(9.8)T^2

Rearranging the above,

1/2)(9.8)T^2 + 8(sin 20)T - 10 = 0

Solving for "T" using the quadratic formula,

T = 0.22 sec.