A long jumper travels 8.95 meters during a jump. He moves at 10.8 m/s when starts his leap. At what angle from the horizontal must he have been moving when he started his jump? You may need the double-angle formula: 2 sin u cos u = sin (2u)
please can someone help?
The formula they are refering to is
X = (V^2/g) sin 2A
8.95 = 10.8^2/9.8 * sin@A = 11.9 sin2A
sin2A = 0.752
2A = 48.8 degrees or 131.2
A = 24.4 or 65.6 deg
The smaller angle is more likely because it allows the long jumper to get added distance by extending feet forward at landing
that is indded correct! can you help me on the other question i had too? its right above this post. it is either the electron or cannon question. thanks so much again!
To find the angle from the horizontal at which the long jumper must have been moving when he started his jump, we can use the equation for horizontal distance:
Horizontal Distance = Initial Velocity * Time * Cos(Angle)
In this case, we are given the horizontal distance (8.95m) and the initial velocity (10.8 m/s). Let's assume the time taken for the jump is 1 second, just for simplicity. So the equation becomes:
8.95m = 10.8 m/s * 1s * Cos(Angle)
Now we need to solve for the angle. Rearranging the equation:
Cos(Angle) = 8.95m / (10.8 m/s * 1s)
Cos(Angle) = 0.827
To find the angle, we need to take the inverse cosine (or arccos) of 0.827:
Angle = arccos(0.827)
Using a calculator, we find that the angle is approximately 33.3 degrees.
However, it is worth noting that the given double-angle formula (2 sin u cos u = sin (2u)) is not directly applicable to solving this problem. It is used for trigonometric identities involving angle addition or subtraction, not for finding angles in projectile motion problems. In this case, we can solve the problem using simple trigonometry without the need for the double-angle formula.