# Projectiles Launched At An Angle

Wayne Brian threw a spear a distance of 201.24 meters. Suppose he threw the spear at a 35.0 degree angle with respect to the horizontal. What was the initial speed of the spear?

Data Known-
Delta X: 201.24 m
Theta (angle): 35.0 degrees
Acceleration (Y): -9.81 meters/seconds/seconds

Equations-
Velocity(Vx)=Vi(Cosine Theta)
Velocity(Vy)=Vi(Sine Theta)
Time= (Delta X)/Vi(Cosine Theta)
Delta Y= (Vi(Sine Theta)*(Time)) + (1/2(Acceleration Y)*(Time^2))
Delta X= Vi(Cosine Theta)*(2*(Vi*Sine Theta)/(Accelartion Y))

How would you solve the problem when all equations require the intial velocity, and there is none given?

cepheid
Staff Emeritus
Gold Member
Welcome to PF!

You know that (Δx)/(vicosθ) is the time required for the projectile to go up and then come back down again. (If it stayed in the air longer, it would have moved farther horizontally. The time spent in the air, along with the horizontal velocity, determines Δx).

You also know that this time spent in the air (t) is equal to twice the time required for the initial vertical velocity (visinθ) to decrease to zero. In other words, the time spent in the air is twice the time spent reaching the maximum height.

So, now you have two equations, and two unknowns (t and vi). You can therefore solve the problem.

Thanks for your response, but I was told by a friend that I could use this equation...

Delta X= Vi(Cosine Theta)*(2(Vi*Sine Theta)/(Acceleration Y))

He said that I would just simplify this equation and plug in the values that I have given to me. Then he said I would just solve for Vi.

So, it ends up looking like this...

201.24 = ((2Vi^2)*(Cos 35)(Sin 35))/(9.81)
(201.24*9.81)/(Cos 35)(Sin 35) = 2Vi^2
Square Root ((201.24*9.81)/2(Cos 35)(Sin 35)) = Vi
Vi= 50.5 meters/seconds

Does this look right?

cepheid
Staff Emeritus