from the roof of atall building, a student throws a ball of mass m with intital speed v=25m/s at an angle = 20 below the horizontal as shown in the figure. the ball is thrown from a height H above the groung. the student measures the ball to be in the air for 4s before it strikes the groung below.
Q1. How far away from the building, D, does the ball land?
Q2. What is the height H from where the ball was thrown?
physic
2007-09-30 4:21 am
回答 (3)
2007-09-30 9:42 am
✔ 最佳答案
Initial velocity of ball u = 25 m/s
vertical component of the velocity of ball
= 25 sin20 (pointing DOWNWARD!!)
horizontal component of the velocity of ball
= 25 cos20
Q1.
its net force in horizontal directoin is ZERO,
so it moves with constant velcity 25 cos20
therefore,
v = D/t
25 cos20 = D/4
D = 93.97m
Q2.
Vertically , it accelerates by g = 10 (pointing DOWNWARD)
by s = ut + (1/2)at^2
s = (4) (-25 sin20) + (1/2) x (-10) x (4)^2 [take downwards as -ve]
s = -34.2 - 80
s = -114.2 m
Thus, the height H = 114.2 m (since height is a scalar quantity, no -ve sign is required.)
p.s. The vertical motion always independent of the horizontal motion, so always resolve the vector quantity (usually velocity) into the vertical and horizontal components and consider these two components independently.
2007-09-30 4:55 am
tan (angle)=x^2 / y^2
so tan(20)=x^2 / y^2
and v^2=x^2+y^2
that tan(20)=v^2-y^2 / y^2
y^2(tan(20)+1)=v^2
y=v / (tan(20)+1)^0.5
H=y*t^2
H=25*16 / (tan(20)+1)^0.5=400 / (tan(20)+1)^0.5.................do it yourself
S=x*t
x^2=v^2-y^2=v^2-v^2 / (tan(20)+1)=v^2 * tan(20) / (tan(20)+1)
x=v * tan(20)^0.5 / (tan(20)+1)^0.5
S=25*4 * tan(20)^0.5 / (tan(20)+1)^0.5=100 * tan(20)^0.5 / (tan(20)+1)^0.5
(do it yourself too)
so tan(20)=x^2 / y^2
and v^2=x^2+y^2
that tan(20)=v^2-y^2 / y^2
y^2(tan(20)+1)=v^2
y=v / (tan(20)+1)^0.5
H=y*t^2
H=25*16 / (tan(20)+1)^0.5=400 / (tan(20)+1)^0.5.................do it yourself
S=x*t
x^2=v^2-y^2=v^2-v^2 / (tan(20)+1)=v^2 * tan(20) / (tan(20)+1)
x=v * tan(20)^0.5 / (tan(20)+1)^0.5
S=25*4 * tan(20)^0.5 / (tan(20)+1)^0.5=100 * tan(20)^0.5 / (tan(20)+1)^0.5
(do it yourself too)
2007-09-30 4:43 am
Initial velocity of ball = 25
vertical velocity of ball = 25 sin 20
horizontal velocity of ball = 25 cos 20
Q1 ,
Horizontally , it is a contsant speed problem
horizontal distance = horizontal velocity x time
distance = 25 cos 20 x 4
distance = 93.97(m)
Q2,
Vertically , it is decelerate by g = -10
by s = ut + 0.5 at^2
s = (4) 25 sin 20 + 0.5 (-10) (4)^2
s = 34.2 - 80
s = -45.8 (m) ................ (negative sign indicate it is downward distance)
so the height H = 45.8 (m)
vertical velocity of ball = 25 sin 20
horizontal velocity of ball = 25 cos 20
Q1 ,
Horizontally , it is a contsant speed problem
horizontal distance = horizontal velocity x time
distance = 25 cos 20 x 4
distance = 93.97(m)
Q2,
Vertically , it is decelerate by g = -10
by s = ut + 0.5 at^2
s = (4) 25 sin 20 + 0.5 (-10) (4)^2
s = 34.2 - 80
s = -45.8 (m) ................ (negative sign indicate it is downward distance)
so the height H = 45.8 (m)
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