A ball thrown vertically returns to the thrower after 6s
Find
(A)the velocity with whixh it was theown up
(B) the maximum height it reaches, and
(C)its position after 4s
Answers
Explanation:
a).initial velocity (u) = 29.4 m/s.
b). the maximum height it reaches is 44.1 metres.
c). 39.2 Metres
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\huge\underline\mathfrak\blue{Explanation}
Total time = 6 sec
Time of ascent = Time of descent = Total time/2 => 6/2 = 3 sec
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a). Given :
final velocity (v) = 0 m/s
acceleration due to gravity (g)= -9.8m/s²
time taken (t) = 3 sec
To find :
initial velocity ( u)
Solution :
We know that,
v = u + at [ 1st equation of motion ]
or, v = u + gt
Put the given values,
0 = u + (-9.8) × 3
0 = u - 29.4
29.4 = u
Therefore, initial velocity (u) = 29.4 m/s. ....(i)
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b). Given :
u = 29.4 m/s [ from (i) ]
g = -9.8 m/s²
t = 3 sec
To find :
the maximum height it reaches.
Solution :
We know that,
S = ut + 1/2 at² [ second equation of motion ]
or, h = ut + 1/2gt²
put the given values,
h = 29.4 × 3 + 1/2 × (-9.8) × 3²
h = 88.2 - 44.1
h = 44.1 m
Therefore, the maximum height it reaches is 44.1 metres.
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c). Given :
u = 29.4 m/s
t = 4 sec
g = -9.8 m/s²
To find :
The position of ball after 4 sec.
Solution :
We know that,
S = ut + 1/2at² [ second equation of motion ]
Or, h = ut + 1/2gt²
h = 29.4 × 4 + 1/2 ( -9.8)(4)²
h = 117.6 - 78.4
h = 39.2 m
Answer:
Explanation:
The ball returns to the ground after 6 seconds.
Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s
Let the velocity with which it is thrown up be u
(a). For upward motion,
v=u+at
∴ 0=u+(−10)×3
⟹u=30 m/s
(b). The maximum height reached by the ball
h=ut+ 1 /2 at2
h=30×3+ 1/2 (−10)×3 2
h=45 m
(c). After 3 second, it starts to fall down.
Let the distance by which it fall in 1 s be d
d=0+ 1/2 at 2 ′
where t =1 s
′
d= 1/2×10×(1) 2
=5 m
∴ Its height above the ground, h
′
=45−5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.