Question

please help me to solve both questions​

Answer 1

Answer:

Range max= Height max/2

Answer 2

45) height the angle theta for the projection should be 45 and the range R is the maximum

h height= u^2sin^2 theta / g

= u^2 /2g...(ans)

. since, sin^2 45 =1/2

46) the initial velocity which is horizontal is 4 metre per second

take an example of a ball which is thrown horizontally from a building top ..

let's take the horizontal Axis-X

and vertical Axis-y

now imagine it to be a part of vectors chapter...

let's make two different words of X and y...

(X)

initial velocity = 4 metre/second

acceleration= 0

time= 0.7 sec

follow the expression,

v=u +at

final velocity in the X axis is

4 m/s

(Y)

initial velocity = 0 m/s

acceleration = 10m/s^2

time = 0.7sec

v= u+ at

final velocity in y axis is 7 m/s

now assume the velocities in x and y axis to be different vectors and we need to find the resultant the resultant will be the velocity at 0.7 seconds which is our answer...

(use triangle vector theorem)

velocity at 0.7 second =v(X) + v(Y)=11m/s ...(ans)

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