Question

. A body is projected horizontally with speed 5 m/s from height h, takes 8 s to reach the ground. The time it takes to cover the first one fourth of height is (1) 4s (2) 6 s (3) 2s (4) 5 s


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Answer 1

Answer:

Given: Speed = 5m/s

           Height = h

           Time = 8 seconds  

To find: Time is taken to cover the first one-fourth of the height

Solution:

Distance traveled by the body when it was thrown with a velocity of 5 m/s

The final velocity will be zero because the body is projected horizontally.

From the law of kinetics,

S = ut + 1/2gt²

 = 5×8 + 1/2×10×(8)²

 = 40 + 320

 = 360 m

According to the question we have to find the time taken by the body to reach one-fourth of height,

One-fourth of height = 360/4 = 80m

Therefore, from the law of kinetics,

S = 1/2gt²

80 = 1/2×10t²

16 = t²

t = 4 seconds

Therefore, the time it takes to cover the first one-fourth of height is 4 seconds.

Answer 2

Answer:

The projectile takes four seconds to cover first one fourth of the height 'h'.

Therefore, option (1) is correct.

Explanation:

We have given, the body projected horizontally with speed, v = 5m/s

The time is taken to reach height h, t = 8s

From the second equation of motion,

$S= ut+\frac{1}{2}gt^2$

The initial velocity of projectile, u = 0

The acceleration due to gravity, g = 10m/s²

Substitute the values of S, t and 'g' in the equation (1);

$h= 0+\frac{1}{2}(10)(8)^2$

$h=\frac{640}{2}m$

$h= 320m$

Consider that t' is the time taken by the projectile to cover one fourth of height 'h',

$\frac{h}{4} =0+\frac{1}{2}(10)(t')^2$

$\frac{320}{4}=\frac{1}{2}(10)t'^2$

$t'^2=16$

$t'=4s$

Therefore, the projectile will take four seconds to cover one fourth height.

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