Question

A projectile cross
half it maximum height at a certain time and again 10 second later. Calculate the maximum height. If the angle of projection was 30 degree, Calculate the maximum range of projectile as well as horizontal distance it travelled in the above 10 second​

Answer 1

Explanation:

motion or two dimensional motion of constant acceleration such as the motion of constant acceleration such as the motion of a particle projected at certain angle with the horizontal in vertical x-y plane (this type of motion is called projectile motion). Air resistance to the motion of the body is to be assumed absent in this type of motion.

A body projected into the space and is no longer being propelled by fuel is called a projectile.

To analyze the projectile motion we use the following concept "Resolution of two dimensional motion into two one dimension motion" as discussed earlier. Hence it is easier to analyze the motion of projectile as composed of two simultaneous rectilinear motions which are independent of each other:

(a) Along the vertical y-axis with a uniform downward acceleration 'g' and

(b) Along the horizontal x-axis with a uniform velocity forward.

Consider a particle projected with an initial velocity u at an angle θ with the horizontal x-axis as shown in figure shown below. Velocity and accelerations can be resolved into two components:

A Particle Projected With an Initial Velocity u at an Angle Theta

Velocity along x-axis = ux = u cos θ

Acceleration along x-axis ax = 0

Velocity along y-axis = uy = u sin θ

Acceleration along y-axis ay = -g

Here we use different equation of motions of one dimension derived earlier to get the different parameters.

\vec{v}=\vec{v_{0}}-\vec{g}t …... (a)

\vec{y}-\vec{y_{0}}=\vec{v_{0}}t-\frac{1}{2}\vec{g}t^{2} …... (b)

v2 = v02 – 2g (y-y0) …... (c)

Total Time of Flight

When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b.

Therefore, 0 = (u sin θ) t - (½)gt2,

As t cannot equal to zero, then, total time of flight,

Horizontal Range

Projectile Motion

Horizontal Range (OA=X) = Horizontal velocity × Time of flight

= u cos θ × 2 u sin θ/g

So horizontal range,

Maximum Height

At the highest point of the trajectory, vertical component of velocity is zero.

Therefore, 0 = (u sin θ)2 - 2g Hmax

So, maximum height would be

2 sin

Hope its help ...

Answer 2

Explanation:

The horizontal distance travelled by a projectile from its initial position (x=y=0) to the position where it passes y=0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight T

f

. therefore, the range R is

R=(v

o

cosθ

o

)(T

f

)=(v

o

cosθ

o

)(2v

o

sinθ

o

)/g

Or, R=

g

v

o

2

sin2θ

o

Equation 2 shows that for a given projectile velocity v

o

, R is maximum when sin2θ

o

is maximum, i.e. when θ

o

=45

o

.

The maximum horizontal range is, therefore