Home
Class 12
PHYSICS
At what angle with the horizontal should...

At what angle with the horizontal should a ball be thrown so that the range `R` is related to the time of flight as `R = 5 T^2` ? `(Take g = 10 ms6-2)`.

A

`45^(@)`

B

`15^(@)`

C

`60^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(2 u^(2) sin theta cos theta)/(g)=5 ((2u sin theta)/(g))^(2)`
`theta = 45^(@)`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEM

    RESONANCE ENGLISH|Exercise DPP No.16|9 Videos

  • DAILY PRACTICE PROBLEM

    RESONANCE ENGLISH|Exercise DPP No.17|20 Videos

  • DAILY PRACTICE PROBLEM

    RESONANCE ENGLISH|Exercise DPP No.14|9 Videos

  • CURRENT ELECTRICITY

    RESONANCE ENGLISH|Exercise High Level Problems (HIP)|19 Videos

  • ELECTRO MAGNETIC WAVES

    RESONANCE ENGLISH|Exercise Exercise 3|27 Videos

Similar Questions

Explore conceptually related problems

A body is projected at an angle of 30^@ with the horizontal and with a speed of 30 ms^-1 . What is the angle with the horizontal after 1.5 s ? (g = 10 ms^-2) .

A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 kmh^(-1) . At what angle of sight (w.r.t horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target? (Take g=10ms^(-2))

A particle is projected from the ground with velocity u at angle theta with horizontal. The horizontal range, maximum height and time of flight are R,H and T respectively. They are given by, R = (u^(2)sin 2 theta)/(g),H = (u^(2)sin^(2)theta)/(2g) and T = (2usin theta)/(g) Now keeping u as fixed, theta is varied from 30^(@) to 60^(@) . Then,

A ball is thrown with velocity 8 ms^(-1) making an angle 60° with the horizontal. Its velocity will be perpendicular to the direction of initial velocity of projection after a time of (g =10ms^(-2) )

A ball is projected with a velocity 20 sqrt(3) ms^(-1) at angle 60^(@) to the horizontal. The time interval after which the velocity vector will make an angle 30^(@) to the horizontal is (Take, g = 10 ms^(-2))

A particle is projected with a speed 10sqrt(2) ms^(-1) and at an angle 45^(@) with the horizontal. The rate of change of speed with respect to time at t = 1 s is (g = 10 ms^(-2))

A ball is thrown at a speed of 40 m/s at an angle of 60^0 with the horizontal. Find a. the maximum height reached and b. the range of te ball. Take g=10 m/s^2 .

The ceiling of a tunnel is 5 m high. What is the maximum horizontal distance that a ball thrown with a speed of 20 ms^(-1) can go without hitting the ceiling of the tunnel (g= 10ms^(-2))

Two projectiles A and B are projected with same speed at an angle 30^(@) and 60^(@) to the horizontal, then which of the following is not valid where T is total time of flight, H is max height and R is horizontal range.

A ball is thrown upwards from the ground with an initial speed of u. The ball is at height of 80m at two times, the time interval being 6 s. Find u. Take g=10 m//s^2.

Home
Profile