You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Summary for Time, Height and Range

Figure shows ...

Question

Figure shows a projectile thrown with speed u= $20$ m/s at an angle $30$ with horizontal from the top of a building $40$ m high then the horizontal range of projectile is.

Open in App

Solution

U = 20 m/s. Ф = 30°. To find x when y = - 40 m.

Equation of projectile: y = x tan Ф - g x² Sec² Ф / 2u²

y = -40 m when the projectile hits the ground.

y = - 40 = x /√3 - (10 * 4/3 * 1/2 * 1/20² ) x²

x² - 20√3 x - 2400 = 0

x = 10√3 + √2700 = 40 √3 meters

Range = 40 √3 m

Suggest Corrections

Similar questions

Q. Figure shows a projectile thrown with speed u = 20 m/s at an angle

30^{0}

with horizontal from the top of a building 40 m high. Then the horizontal range of projectile is

Q. A projectile thrown with an initial speed

u

and the angle of projection

15^{o}

to the horizontal has a range

R

. If the same projectile is thrown at an angle of

45^{o}

to the horizontal with speed

2 u

, what will be its range?

Q. A projectile is fired with an initial speed of

5 m/s

at an angle of

45^{\circ}

with the horizontal, up an inclined plane having angle of inclination

30^{\circ}

from horizontal. What is the range of this projectile?
(Take

g = 10 m / s^{2}

)

Q. A projectile is fired horizontally with a velocity of

98 m / s

from the top of a hill

490 m

high. Find the horizontal range of the projectile.

A projectile thrown with an initial speed u and angle of projection $15^{\circ}$ to the horizontal has a range R. If the same projectile is thrown at an angle of $45^{\circ}$ to the horizontal with speed 2u, its range will be

Join BYJU'S Learning Program

Figure shows a projectile thrown with speed u=20 m/s at an angle 30 with horizontal from the top of a building 40 m high then the horizontal range of projectile is.

U = 20 m/s. Ф = 30°. To find x when y = - 40 m. Equation of projectile: y = x tan Ф - g x² Sec² Ф / 2u² y = -40 m when the projectile hits the ground. y = - 40 = x /√3 - (10 * 4/3 * 1/2 * 1/20² ) x² x² - 20√3 x - 2400 = 0 x = 10√3 + √2700 = 40 √3 meters Range = 40 √3 m

Figure shows a projectile thrown with speed u= $20$ m/s at an angle $30$ with horizontal from the top of a building $40$ m high then the horizontal range of projectile is.

U = 20 m/s. Ф = 30°. To find x when y = - 40 m.

Equation of projectile: y = x tan Ф - g x² Sec² Ф / 2u²

y = -40 m when the projectile hits the ground.

y = - 40 = x /√3 - (10 * 4/3 * 1/2 * 1/20² ) x²

x² - 20√3 x - 2400 = 0

x = 10√3 + √2700 = 40 √3 meters

Range = 40 √3 m