If an object is projected vertically upward with a velocity of 50 m/s, and the time of flight is given as 60 seconds, the value of tangent you are asking about is unclear since the term "tangent" typically relates to trigonometry (the ratio of the opposite side over the adjacent side in a right triangle).
However, if we assume you're talking about the time of flight and the initial velocity of a vertically launched projectile, we can calculate the maximum height and the acceleration due to gravity that would result in a 60-second total flight time.
The time of flight for an object projected vertically with initial velocity \(v_0\) and then returning to the ground is given by:
\[ t = \frac{2v_0}{g} \]
where \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \) on the surface of the Earth.
Plugging in the numbers:
\[ 60 \, \text{s} = \frac{2 \times 50 \, \text{m/s}}{g} \]
Solving for \( g \):
\[ g = \frac{2 \times 50 \, \text{m/s}}{60 \, \text{s}} \]
\[ g = \frac{100 \, \text{m/s}}{60 \, \text{s}} \]
\[ g \approx 1.67 \, \text{m/s}^2 \]
This calculated value for \( g \) is much less than the actual acceleration due to gravity on Earth, and a 60-second time of flight for an object thrown vertically with an initial velocity of 50 m/s is not possible under normal Earth gravity.
With the actual acceleration due to gravity:
\[ t = \frac{2v_0}{g} \]
\[ t = \frac{2 \times 50 \, \text{m/s}}{9.81 \, \text{m/s}^2} \]
\[ t \approx \frac{100 \, \text{m/s}}{9.81 \, \text{m/s}^2} \]
\[ t \approx 10.19 \, \text{s} \]
This would be the total time of flight for an object thrown vertically upwards with a speed of 50 m/s under normal Earth gravity, which is significantly less than 60 seconds.
If you meant something different with the term "value of tangent," please provide additional context or clarify the question so that I can assist you accordingly.
An object is projected with a volecity of 50m/s at ground vertical. If the total time of fight of project is 60s. What is the value of tangent
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