Calculate the time of fight for a horizontally launched projectile from a height of 20m above the ground with an initial velocity of 5m/s. Please

0.5sec

Explanation:

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Method one

Using the second law of motion

H=ut-1/2gt^2

20=5t-1/2×10×t^2

20=5t-5t^2

dh/dt = 5-10t

where any constant is zero therefore the 20 is zero

5-10t=0

Collect like terms

-10t= -5

t=1/2 = 0.5sec

2nd method

Parameters

Height(H) =20m

Initial velocity(u) = 5m/s

Acceleration due to gravity(g) =10m/s^2

Using the time taken formula

t=u/g

t=5/10

t=0.5sec

the answer is a i think

+15-35=-20

going in opposite directions