Find time of flight and range of the projectile along the inclined plane as shown in figure. $(g = 10 m / s^{2})$

T = \frac{2}{\sqrt{3}} s, R = \frac{80}{3}

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T = \frac{4}{\sqrt{3}} s, R = \frac{160}{3} m

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T = \frac{8}{\sqrt{3}} s, R = \frac{160}{3} m

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T = \frac{2}{\sqrt{3}} s, R = \frac{160}{3} m

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Solution

The correct option is A $T = \frac{4}{\sqrt{3}} s, R = \frac{160}{3} m$
$L e t t h e x a x i s b e a l o n g t h e i n c l i n e a n d y a x i s b e p e r p e n d i c u l a r t o i t . u_{x} = 20 c o s 30^{0} = 20 \frac{\sqrt{3}}{2} = 10 \sqrt{3}, a_{x} = 10 s i n 30^{0} = 5 u_{y} = 20 s i n 30^{0} = 20 \frac{1}{2} = 10, a_{y} = - 10 c o s 30^{0} = - 5 \sqrt{3} Y = 0 = u_{y} t - \frac{1}{2} a_{y} t^{2} = 10 t - \frac{1}{2} 5 \sqrt{3} t^{2} \Rightarrow T = \frac{4}{\sqrt{3}} s R = u_{x} T + \frac{1}{2} a_{x} T^{2} = 10 \sqrt{3} \times \frac{4}{\sqrt{3}} + \frac{1}{2} \times 5 \times {(\frac{4}{\sqrt{3}})}^{2} R = \frac{160}{3} m$

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Similar questions

Q. A body is thrown with a velocity of

20 m s^{- 1}

from a building of height

15 m

. If the angle of projection is

30^{\circ}

above the horizontal, the time of flight and the range of the projectile are respectively.
(Take

g = 10 m / s^{2}

)

Q. Find time of flight and range of the projectile along the inclined plane as shown in figure

g = 10 m / s^{2}

Q. Find the cartesian form of the equation of the plane

¯ r . [(s + t)^i + (2 + t)^j + (3 s + 2 t)^k] = 15

where

s

and

t

are parameters

When a particle is projected at an angle to the horizontal, it has range R and time of flight t1. If the same projectile is projected with the same speed at another angle to have the same range and time of flight $t_{2}$ then,

The position of a body moving along x-axis at t is given by x=(t²-4t+6)m. The distance travelled by body in time interval t=0 to t=3 s is
(A) 5m
(B) 7m
(C) 4m
(D) 3m

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Find time of flight and range of the projectile along the inclined plane as shown in figure. (g=10m/s2)

The correct option is A T=4√3s,R=1603mLetthexaxisbealongtheinclineandyaxisbeperpendiculartoit.ux=20cos300=20√32=10√3,ax=10sin300=5uy=20sin300=2012=10,ay=−10cos300=−5√3Y=0=uyt−12ayt2=10t−125√3t2⇒T=4√3sR=uxT+12axT2=10√3×4√3+12×5×(4√3)2R=1603m

Find time of flight and range of the projectile along the inclined plane as shown in figure. $(g = 10 m / s^{2})$