A projectile was launched from the ground. It had a range of 70 metres and was in the air for 3.5 seconds.
At what angle to the horizontal was it launched?
Hey there, welcome to the forums!! A bit of a mean question this one, they don't give much away
Let's consider the horizontal component of velocity first. We know the range and time of flight, so it is actually a straightforward equation:
For vertical, we need to find the velocity required for the vertical displacement to be zero at t=3.5:
Now we form a right angled triangle with these values, and the angle we want is just the inverse tangent of the vertical and horizontal velocities:
This last bit might be a bit hard to picture, but remember that we can break a launch velocity into its vertical and horizontal components! This forms a right angled triangle, with the vertical side corresponding to vertical velocity, and the base to horizontal velocity
let me know it this makes sense!