A body is projected by making an angle 30 degree with the horizontal with a velocity of 39.2 m/s ,find
a)Time of flight(T)
b)Range (R)
c)Maximum height (H)
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A body is projected by making an angle 30 degree with the horizontal with a velocity of 39.2 m/s ,find
a)Time of flight(T)
b)Range (R)
c)Maximum height (H)
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Step-by-step explanation:
Maximum height,
H = u² sin²β/2g = 30²× sin²30°/(2×10) = 11.25 m
Time of flight ,
T = 2u sinβ /g = 2×30 ×sin30°/10 =3 sec
Horizontal range,
R= u²sin2β/g = 30² × sin60°/10 = 45√3 m
(a). The time of flight is 2 sec.
(b). The range is 135.79 m.
(c). The maximum height is 19.6 m.
Step-by-step explanation:
Given that,
Angle = 30°
Horizontal velocity = 39.2 m/s
(a). We need to calculate the time of flight
Using formula of time of flight
Put the value into the formula
(b). We need to calculate the range
Using formula of range
Put the value into the formula
(c). We need to calculate the maximum height
Using formula of maximum height
Put the value into the formula
Hence, (a). The time of flight is 2 sec.
(b). The range is 135.79 m.
(c). The maximum height is 19.6 m.
Learn more :
Topic : projectile motion
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