Question

0. A ball is tossed from an upper-story window of a bui lding. The ball is given an initial velocity of 9.30 m/s at an angle of 22.00 below the horizontal. It strikes the ground 4.o0 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? My Notes Ask Your -/1 points SerPSE9 4.P037. ii a horiznntal, lindrical tybe 58.0 ft lonq and is represented in the figure below. Assume an astronaut in training sits in a seat at one

Answer 1

Solution :-

initial velocity= 9.3 m/sec

angle =22⁰

so the horizental velocity

$v_{x}=vcos\Theta$

= 8.62 m/s

a)

so horizental distance covered by the ball is

H=v_xt=8.62×4 =34.48 m

so the answer is 34.48 m

b)vertical velocity

$v_{y}=vsin\Theta$

= 8.62sin22

= 3.23 m/s

$s=v_{y}t+\frac{1}{2}gt^{2}$

= 3.23×4 + (0.5)×9.8×(4)^2

= 91.32 m

in 3 significant figure the height is 91.3 m

-------------------------------------------------------------

c)

$v^{2}-v_{y}^{2}=2gh$

$v^{2}=2gh+v_{y}^{2}$

$v=\sqrt{2gh+v_{y}^{2} }$

= 14.37 m/s

$v-v_{y}=gt$

$t=\frac{v-v_{y}}{g}$

= (14.37- 3.23)/(9.8)

= 1.1377 sec

in three significant figure

t = 1.14 sec