Question

For a projectile thrown with a speed 50 ms-1 at an angle 37° with the horizontal, the time of flight will be [g = 10 m/s2]?​

Answer 1

Answer:

The time taken by the projectile to reach the ground is 10sec.

The horizontal range is 500m.

Initital velocity = 50m/s , in horizontal direction

Initial height = 500m

g= 10m/s^2

By equations of motion, s =ut + 0.5at^2, s = distance,a = acceleration

u = initial velocity in vertical direction = 0m/s

by equation, s = 0.5at^2 ==> t = \sqrt{\frac{2s}{a} } = \sqrt{\frac{2h}{g} }

a

2s

=

g

2h

t = \sqrt{\frac{1000}{10} } = \sqrt{\frac{100}{1} } =10s

10

1000

=

1

100

=10s

To find the horizontal range,

Here acceleration in horizontal direction = 0.

Using same equation as before, R =Vt + 0.5at^2

Range = Vxt, V = velocity in horizontal direction, a constant since a =0

Range = 50m/s x 10s = 500m

Answer 2

The time of flight of the projectile that is thrown is 6 seconds.

Given,

Projectile thrown with speed=50 m/sec

The angle at which it is thrown=37° with the horizontal.

To find,

the time of flight of the projectile.

Solution:

• When motion is a projectile motion, the time of flight is given as:
• $T=\frac{2usin(angle)}{g}$.
• where, u-initial velocity, angle=angle with the horizontal and g-acceleration due to gravity.
• sin 37°=3/5.

The time of fight for the entire journey will be:

$T=\frac{2usin(angle)}{g}\\T=\frac{2X50X\frac{3}{5} }{10} \\T=\frac{2x10X3}{10} \\T=2X3\\T=6 seconds.$

Hence, the time of flight is 6 seconds.

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