Question

A ball is lossed from an upper-story window of a building. The ball is given an initial velocity of 7.60 m's at an arngle of 24.0° below the horizontal. Il strikes the grournd 5.00 s later (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown (C) How long does it take the ball to reach a point 10.0 m below the level of launching?

Answer 1

(a) We know that
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 = 7.60(cos24)(5)

X = 34.69 m

b) We know that
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 = 7.60*(sin 24)(5) + (1/2)(9.8)(5^2)

Y = 137.92 meters

c) Formula is

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

where

S = 10 m
Vy = 7.60*(sin 24)
g = 9.8 m/sec^2
T = time for ball to reach a point 10 m below the launching level

Substituting values,

10 = 7.60*(sin 24)T + (1/2)(9.8)T^2

Rearranging the above,

0.5*(9.8)T^2 + 7.60*(sin 24)T - 10 = 0
4.9T^2 + 3.085T-10=0
Solving for "T" using the quadratic formula,

T = 1.147 sec.