1. Rahim, the centre on the basketball team is 1 7/10 m tall. His rival for the position is 1 5/8
m tall. What is (a) there total height ,(b) the difference between their height
Answers
Length of a room =9m,
Breadth of a room = 8m
And height of room = 6.5 m
∴ Area of 4 walls = Lateral surface area
= 2 (l+ b) x h
= [2 (9+8) x 6.5] m2
= (2 x 17 x 6.5) m2
=221 m2
∴ Area not be whitewashed = (area of 1 door) + (area of 2 windows)
= (2 x 1.5) m2 + (2 x 1.5 x 1) m2
= 3m2 + 3m2 =6m2
∴ Area to be whitewashed = (221-6) m2 =215 m2
∴ Cost of whitewashing the walls at the rate of Rs.6.40 per
Square meter = Rs. (6.40 x 215) = Rs. 1376
Step-by-step explanation:
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Class 11
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>>Motion in a Plane
>>Projectile Motion
>>A basketball player is standing on the f
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A basketball player is standing on the floor 10.0m from the basket as in above figure. The height of the basket is 3.05m, and he shoots the ball at a 40.0
0
angle with the horizontal from a height of 2.00m. (a) What is the acceleration of the basketball at the highest point in its trajectory? (b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
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Updated on : 2022-09-05
Solution
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Verified by Toppr
a) The acceleration is that of gravity: 9.80m/s
2
, downward.
(b) The horizontal component of the initial velocity is v
xi
=v
i
cos40.0
0
=0.766v
i
, and the time required for the ball to move 10.0m horizontally is
t=
v
xi
Δx
=
0.766v
i
10.0m
=
v
i
13.1m
At this time, the vertical displacement of the ball must be
Δy=y
f
−y
i
=3.05m−2.00m=1.05m
Thus, Δy=v
yi
t+
2
1
a
y
t
2
becomes
1.05m(
v
i
sin40.0
0
)
v
i
13.1m
+
2
1
(−9.80m/s
2
)
v
i
2
(13.1m)
2
or 1.05m=8.39m−
v
i
2
835m
3
/s
2
which yields,
v
i
=
8.39m−1.05m
835m
3
/s
2
=10.7m/s