Kinematic formula for projectiles

[answerseeker]

Q: an object is thrown from the ground into the air with a velocity of 20.0m/s at an angle of 27.0 degrees to the horizontal. what is the max height reached by this object?

i drew the triangle and found vertical side to be 9.08 m/s and horizontal side to be 17.8m/s using sin and cos.
/|
20 / | 9.08
/__|
17.8

but the part I'm having trouble with is how to do the horizontal part of the equation using d=vt. I am assuming that i have to use that eqn to find time so i can use that value for the vertical part in finding d, using kinematic formula.
in the horizontal part d=vt, what would i use as d? i have v as 17.8

also a side note: for other projectile questions, when would you designate final velocity as unknown, zero, or the negative sign of the initial velocity (ie. initial velocity=14, final velocity = -14) ?

 

[whozum]

The key to this problem is to find the time of flight by knowing the initial vertical velocity. From there you'll be able to find time of flight, horizontal distance, vertical distance.

 

[quicknote]

The kinematic formula for projectiles can be used to calculate the maximum height reached by an object thrown at an angle. In this case, the initial velocity (vi) is 20.0m/s and the angle (θ) is 27.0 degrees. The vertical component of the initial velocity can be calculated using the sine function: viy = vi*sin(θ) = 20.0*sin(27.0) = 9.08m/s.

To find the maximum height reached by the object, we can use the kinematic formula for vertical motion: Δy = viyt + (1/2)at^2, where Δy is the displacement (maximum height), viy is the initial vertical velocity, a is the acceleration due to gravity (9.8m/s^2), and t is the time.

We can find t by using the kinematic formula for horizontal motion: Δx = vixt, where Δx is the horizontal displacement, vix is the initial horizontal velocity, and t is the time. In this case, Δx is the distance traveled horizontally, which is equal to the range of the projectile. So we can use the range formula: R = (vi^2 * sin(2θ))/g, where R is the range, vi is the initial velocity, θ is the angle, and g is the acceleration due to gravity.

Substituting in the given values, we get: R = (20.0^2 * sin(54.0))/9.8 = 17.8m. This is the distance traveled horizontally by the object.

Now, we can use the range formula again to find the time: t = R/vix = 17.8/20.0*cos(27.0) = 0.83s.

Finally, we can plug in the values for t and viy into the formula for vertical motion: Δy = 9.08*0.83 + (1/2)*(-9.8)*(0.83)^2 = 3.77m. This is the maximum height reached by the object.

To answer your side note, the designation of final velocity as unknown, zero, or negative of the initial velocity depends on the specific situation and what information is given. Generally, if the object is thrown vertically upward, the final velocity will be zero at the highest point. If the object