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3. A projectile is fired from ground level with an initial velocity of 35 m/s at an angle of 35° with the horizontal. How long will it take for the projectile to reach the ground?

User Jsuggs
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2 Answers

4 votes

Answer:

4.1 seconds to the nearest tenth.

Explanation:

The vertical component of the velocity = 35 sin 4.935 m/s.

The relation between the height (h) of the projectile and time is given by:

h = ut + 1/2 g^2 where u = initial velocity, t = the time and g = acceleration due to gravity which we can take to be 9.8 m/s/s. When the projectile hits the ground h = 0 .

So we have h = 35sin35 t - 4.9t^2 = 0

t(35sin35 - 4.9t) = 0

4.9t = 35 sin35

t = 35 sin 35 / 4.9

= 4.097 seconds

User Spikes
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6.0k points
2 votes

Answer: 4.10s

t=2Vi*sin(theta)/g

Vi=initial velocity=35m/s

g=9.8m/s^2

t=2*35*sin(35)/9.8=4.10s

User Obinna Kalu
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6.1k points